SMALL MOTORS, TRANSFORMERS, 
ELECTROMAGNETS 



A PRACTICAL PRESENTATION OF DESIGN AND CONSTRUCTION 
DATA FOR SMALL MOTORS, SMALL LOW- AND HIGH- 
TENSION TRANSFORMERS, ELECTROMAGNETS, 
AND INDUCTION COILS 



H. M. STOLLER, B.E., M.S. 

' DEVELOPMENT ENGINEERING DEPARTMENT, WESTERN ELECTRIC COMPANY 

F. E. AUSTIN, B.S., E.E. 

PROFESSOR OF ELECTRICAL ENGINEERING AND MECHANICS, THAYER SCHOOL 
OF ENGINEERING, DARTMOUTH COLLEGE, HANOVER, NEW HAMPSHIRE 

E. W. SEEGER, M.E. 

ELECTRICAL ENGINEER, THE CUTLER-HAMMER MANUFACTURING COMPANY 



ILLUSTRATED 



AMERICAN TECHNICAL SOCIETY 

CHICAGO 

1920 



<v<,?r ^ 



COPYRIGHT, 1920, BT 

AMERICAN TECHNICAL SOCIETY 



COPYRIGHTED IN GREAT BRITAIN 
ALL BIGHTS RESERVED 



DEC - 1 1920 



g)CU601768 



INTRODUCTION 



"C^ROM the electrical industry has come a great demand for prac- 
•^ tical information on the design, construction, and operation of 
small motors, small low- and high-tension transformers, electro- 
magnets, and induction coils; information that is not overweighted 
with theory but is yet authoritative. 

^ The three sections of this book will fully meet that demand, 
for they are written in simple, non-technical language; even the 
amateur can easily follow the instructions. At the same time the 
book does not depart in any way from the high standards of engi- 
neering practice; it is valuable ahke to the electrical engineer and 
to the amateur. 

^ The authors of these "three books in one" hold high places in 
the electrical industry either as engineers foremost in their profes- 
sion or as practical successful designers in commercial work. 

^ "Design of Small Motors," the first section of this volume, 
describes with careful attention to detail the practice of the Western 
Electric Company in designing small motors. Typical designs are 
given of direct-current motors of all standard voltages and ranging 
in size from ywo ^P- ^P ^^ i ^P-' ^^^ ^^ alternating-current induc- 
tion motors, ranging in size from ^ to ^ hp. The starting and 
Ughting equipment for automobiles is illustrated by the design of a 
starting motor and a charging generator. The farm-lighting out- 
fit is also fully covered. 

^ One of the most valuable features of this section is the instruc- 
tion given on the rewinding of both direct-current and alternating- 
current motors of small sizes. Practical cases are assumed, such 
as rewinding for changes in voltage and changes in speed and infre- 
quency; rewinding induction motors where the name plate is missing 
and some of the windings have been removed. 

^ "Design of Small Low- and High-Tension Transformers" is 
written to meet the demand which has arisen for more practical 
knowledge of the low-tension transformer which will transform 
from 110 to 220 and down to the lower voltages. This section 
also covers the design and particularly the construction of small 
high-tension transformers, such as would be used for wireless work, 
and full information is to be had as to the design and construction 
of a welding transformer. The author has treated the principles 
of the welding transformer in a manner which can be easily under- 
stood by the non-technical reader as well as the engineer. 



INTRODUCTION 

^ "Design of Electromagnets and Induction Coils" gives, first, 
the modern theory of the magnetic circuit. Then follow typical 
designs of the various types of electromagnets, carrying through 
the calculations to the final result. Of great value to the electrical 
worker are designs of the flat-plunger electromagnet, the cone- 
plunger, the horseshoe and the clapper types, and the portative, 
or holding, magnet, for both direct-current and alternating-current 
work, together with the induction-motor type for alternating 
current. 

^ While it is almost impossible to give all the many applications 
to which electromagnets are put in commercial work, keen prac- 
tical judgment has been exercised by the author as to the appli- 
cations of the electromagnet described in this volume. An added 
feature which will appeal to the reader is the treatment accorded 
the methods of locating trouble in magnets which are in service. 

^ "The Induction Coil, Its Design and Construction," appears 
in the second part of "Design of Electromagnets and Induction 
Coils," and there is included also a concise description of the stand- 
ard Tesla coil. 

^ In offering this volume to the electrical engineer and worker, 
the publishers feel confident it will more than satisfy the demand 
for a book that contains all the essential details of design, construc- 
tion and operation. 



CONTENTS 

Design of Small Motors 



y 

■ PAGE 

iRneral principles 1 

Direct-current motors 2 

Induction motors 12 

Universal motors 21 

Direct-current motor design 23 

•7^5-hp. 110-volt shunt motor 23 

y^5-hp. 12-volt shunt motor 36 

Y^-^hp. series motor 40 

Design of type SDA frames 47 

•|-hp. 110-volt compound motor 62 

Induction motor design 59 

• •J-hp. motor 59 

^hp. motor .- 73 

Design of induction-motor frames 83 

Changing characteristics of motor by rewinding 85 

Direct-current motors 85 

^^Case 1 85 

^BCase II 90 

^BCase III — Change of voltage 96 

^f Case IV — Change of speed 97 

Case V — Change in direction of rotation 99 

Induction motors 99 

Case VI 100 

Case VII — Change of frequency 105 

Case VIII — Change of voltage 110 

Case IX — Change of speed Ill 

Automobile starting and lighting equipment 113 

Starting motor design ". 113 

Starter for engine of 300 cubic inches displacement 117 

Starter for engine of 200 cubic inches displacement 124 

Charging generator design 126 

Farm lighting outfits 135 

Design of Small Low- and High-Tension Transformers 

Theory of transformer operation 146 

Relation between electric current and magnetic field 146 

Magnetic conductivity 148 

Retentivity 149 

Magnetic induction 150 

Electromotive force of induction 150 

Right-hand rule 151 

Equations of induced pressure 152 

Left-hand rule 155 



CONTENTS 

PAGE 

Electromotive force of induction (continued) 

Counter-electromotive force 155 

Rate of change in magnetic flux and current . .-. 157 

Coefficient of inductance 160 

Magnetic leakage 163 

Losses in transformer 165 

Primary and secondary losses . 165 

Core loss 166 

Power efficiency of transformers 168 

Variation in efficiency 169 

All-day efficiency 169 

Regulation of transformer 170 

Design of 10000-volt transformer 171 

Design of core 173 

Design of primary 175 

Design of secondary 178 

Construction of transformer 179 

Miscellaneous details 190 

Design of 2200-volt transformer : 197 

Primary current 199 

Secondary current 203 

Size of core 204 

Length of primary wire 207 

Length of secondary windings 209 

Design of 10-volt transformer 213 

Number of turns in primary coils 213 

Volume of iron core 215 

Length of primary wire 217 

Number of turns on secondary 221 

Connecting coils for different pressures 223 

Design of 22-volt transformer 226 

Volume and weight of core 227 

Number of turns in primary windings 228 

Number of turns in secondary 229 

Design of case for 220-volt transformer 231 

Recapitulation 234 

Core and copper loss . 234 

No-load primary loss 238 

Design of Electromagnets and Induction Coils 

Electromagnets 241 

Theory of magnetic circuit 241 

Magnetomotive force 242 

Reluctance • . . . . 243 

PermeabiUty 244 

Magnetic intensity 246 

Coil construction 249 

Bobbin-wound coils 251 

Form-wound coils 252 



P CONTENTS 

PAGE 

binding calculation 253 

Jirect-current electromagnets 262 

Flat-plunger electromagnet 264 

B Cone-plunger electromagnet 275 

™ Horse-shoe electromagnets 279 

Clapper type electromagnet 282 

Portative, or holding, magnets 283 

\lteniating-current electromagnets 286 

I Single-phase magnets. 289 

Polyphase magnets 290 

Plunger electromagnet 290 

Horseshoe and clapper type electromagnets 292 

Induction motor type 292 

Portative, or holding, magnets 292 

Electromagnet applications 293 

Magnetic separator pulley 294 

Lifting magnet 295 

Magnet switches 297 

Magnetic brakes 299 

Faults in electromagnets 299 

Residual magnetism 299 

Coil troubles 300 

Induction coils 300 

Primary induction coils 300 

Secondary induction coils 302 

Construction of induction coil 304 






RICHMOND POLYPHASE MOTOR DIRECT CONNECTED TO BAND SAW 

Courtesy of Richmond Electric Company 




DESIGN OF SMALL MOTORS 

PART I 



GENERAL PRINCIPLES 

Fundamental Design. In small motors the labor cost is 
far greater than the cost of materials. For this reason it is essen- 
tial that the design should be simple, with as few parts as pos- 
sible. It is necessary that these parts should be interchangeable 
in order to simplify their assembly. Consequently the design 
should be such that small variations in the dimensions of the 
parts will not make the assembly difficult or affect the operation 
of the motor. The shop cost of machine work held to close 
limits is very high, and the number of parts requiring accurate 
work should therefore be confined to a minimum. Manufactur- 
ers usually employ a set of standard frame sizes proportional to 
the power output, and by varying the windings and speeds meet 
various voltage and current requirements. By this means the 
number of different mechanical parts is kept down to a mini- 
mum and the quantity of each part is raised to a maximum, 
thereby reducing the cost of production. 

In determiniijg the electrical features of small motor design 
the procedure followed does nor differ greatly from that employed 
in designing large motors, but the values of current density, flux 
density, speed, etc., which are essential for a practical design, are 
very different from the corresponding values in large motors. 
This information is given in the various illustrative designs which 
are later worked out in detail. It is not possible to predetermine 
the performance of a small motor with as great accuracy as that 
of a large motor, because in the former the losses are relatively 
greater. It is, in fact, necessary to make several trials before a 
given set of requirements can be met exactly, and even an experi- 
enced designer cannot expect the performance of a new design to 
agree closely with the calculated values of efficiency, temperature 
rise, etc. The designer must therefore regard his calculations as 
approximations onlj'. In order, however, to minimize the amount 



2 DESIGN OF SMALL MOTORS 

of ''cut-and-try" work, it is essential to make complete calculations 
so that the first model will not be very far off and so that the 
experimental test may be given the proper interpretation to 
ensure that the second model will be satisfactory. These calcula- 
tions involve the" use of many semi-empirical formulas and rules 
which are the result of experience but which are difficult to dis- 
cuss apart from specific examples. A number of designs of various 
sizes of both d.c. and a.c. motors have been worked out in suffi- 
cient detail to enable each step to be followed. These designs are 
preceded by comments on general mechanical construction. 

Direct=Current Motors.* The dimensions of a line of small 
d.c. motors sold by the Western Electric Company are shown in 
Fig. 1; and the construction and the general appearance of the 
parts are shown in Fig. 2. In Fig. 3 is illustrated a similar line 
of motors manufactured by the Robbins & Myers Company. 
Fig. 4 shows a disassembled view of frame No. 105. It is feasible 
to employ quite large steps between frame sizes if the design is 
arranged so that the length of frame may be varied for a given 
diameter. This is easily accomplished, in case the field is built 
up of punchings, by merely varying the number of punchings. 

Armature. The armature punchings of small d.c. motors are 
almost without exception built for direct assembly on the shaft, thus 
eliminating the armature spider which is employed in large 
machines. Fig. 5 shows a typical punching for a small motor. 
It should be noted that overhung teeth are used, which give a 
small slot opening as compared with the open slots used in large 
motors. It will be shown in the illustrative designs that these 
nearly closed slots are necessary in order to reduce the field 
ampere turns and keep the losses in the field coils within reason- 
able limits. Incidentally the overhung teeth prevent the armature 
coils from being thrown out by centrifugal force. The teeth are 
made wdth parallel sides, thus giving the maximum slot width. 
There is usually sufficient core section beneath the slots to carry 
the flux with reasonable density, so that this does not influence 
the depth of slot. The shaft should not be considered as part 
of the core, since it is solid. The armature windings are similar to 

* See "Design of Direct-Current Dynamos"; and "Management of Dynamo-Electric 
Machinery," Part I. 



DESIGN OF SMALL MOTORS 





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4 DESIGN OF SMALL MOTORS 

those employed on large motors, except that, in general, they con- 
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Fig. 2. Parts of Western Electric Type SD Direct-Current Motor 

account of the nearly closed slots. Short-pitch, or corded wind- 
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DESIGN OF SMALL MOTORS 



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6 DESIGN OF SMALL MOTORS 

resist centrifugal stresses. A cording of one or two slots less thai 
the pole pitch is a favorite winding. After the winding is com- 
plete, the armature is treated with sterling varnish, bakelite, oi 
other compound which acts as a binder as well as a protectior 
against moisture. 

Commutator. Commutator construction is a feature which 
has been given considerable study, and several methods of manu- 
facture have been evolved. A recent practice introduced by the 
Westinghouse Company which is meeting with favor in small 
motor construction is to punch out the bars from stock of tapered 




Fig. 4. Robbins &, Myers Direct-Current Motor, Frame 105, One-Fourth Horsepower, i . 

1650R.P.M. II 



thickness. These are assembled around a steel core drilled to fit 
the shaft and separated by micanite sheets. The entire com- 
mutator is then molded together with bakelite and turned true. 
Another method of commutator construction is to start with a 
solid copper cylinder and form the individual bars by "milling in" 
slots. The bars are then insulated with mica and clamped between 
steel V rings. 

Field Structure. The field structure is made in a number of 
different forms. A favorite design for very small motors employ- 
ing one punching for the entire field is shown in Fig. 6. The 
punchings are assembled in a thin steel tube which forms the outer 



DESIGN OF SMALL MOTORS 



casing and serves as a bearing surface for the end shields. This 
gives extreme simpUcity with a good magnetic circuit. The field 
coils are wound on a form, but as they are not impregnated they 

TYPt-SD, FRAME N9 264- 

COMMUTATOR' ^2" DIA. 
I" LONG. 50BAR6. 



FRAME LENGTH' 2.5" 
AIR GAP .0625" 
NO SLOTS 25 
SLOT DEPTH .50" 
SLOT WIDTH (AVE.) .25" 



SCALE: QUARTER SIZE 

Fig. 5. Western Electric Type SD, Frame 264, Direct-Current Motor 

are flexible and may be forced over the pole tips. A soft-iron strip 
is inserted through the semicircular space back of the pole and 
the ends bent up to retain the field coil. 

Another niethod of field construction (see Fig. 37), employs 
a thick soft-steel tube, the ends of which are turned to seat the 
end shields. The field poles are made either of solid wrought iron 
or of a group of punchings riveted together. The back surfaces 

TYPE'DSA. FRAME N9 305 
LENGTH I" 
AIR GAP, 0125" 





COMMUTATOR: ,„ 
I" DIAMETER. I" LONG. 
22 BARS ^ 

5CALE: HALF SIZE 

Fig. 6. Western Electric Type SDA, Frame 205, Direct-Current Motor 

are accurately machined to fit snugly against the inside of the 
tubular frame. The field coils are form wound and fitted over the 
poles. The pole cores are usually held by means of a single large 



8 V DESIGN OF SMALL MOTORS 

macliine screw from the outside through the frame into a threaded 
hole in the poles. 

Bearings and Brush Rigging. The end brackets are castings, 
usually of iron or aluminum, and contain the bearings and brush 
rigging. In very small motors a favorite bearing construction is 
to employ a thin sleeving of babbitt or similar metal, which is 
lubricated by means of a wick. The wick is gripped within a few 
turns of a small compression spring which fits into a cartridge 
packed with vaseline or similar grease and threaded into the bear- 
ing. In larger designs of i hp. and over an oil pocket is generally 
provided, together with an oil ring dipping into the oil and carry- 
ing it up on the shaft as it revolves. 

Several different types of brush rigging are in use. In one 
of the types employed in the Western Electric motors, Fig. 2,| 
bosses are cast in the end shields. These are drilled and fitted 
with bushings of fiber N. Inside this bushing and integral with! 
it is a square metal sleeve in which the brush slides and to which 
the leads are attached. The brushes J are fitted with com- 
pression springs which bear against cap screws / threaded into 
the fiber bushing. This arrangement gives easy access to the 
brushes and allows their removal without disconnecting any leads, j 
Pigtails are not provided on sizes below J hp., with the exception 
of low- voltage motors, where they are always employed. The pig- 
tails are coiled within the compression spring and soldered to the 
end. The current then enters the brush, flowing through the pig- 
tail and out by way of the cap screw to the metal sleeving and 
inside terminals as shown in Fig. 2. 

In another type of brush holder employed by the Westing- 
house Company in their automobile equipment a micarta ring is 
provided which is slotted and clamped against the inside surface 
of the end shield by means of two machine screws running through 
the end shield with the heads on the outside. The micarta ring 
carries a punched "box" brush holder and a clock spring of either 
steel or phosphor bronze which bears against the end of the brush. 
The pigtail ends in a terminal which is held under a machine 
screw threaded into a lip on the brush box, to which inside leads 
are run. This construction permits easy adjustment of the brush 
position by turning the micarta ring. 



8 



TABLE I 
Comparison of Small and Large D.C. Motor 





i Horsepower 


25 Horsepower* 


Rating 






Horsepower output 


i 


25 


Number of poles 


2 


4 


Voltage 


32 


220 


Speed 


2000 r.p.m. 


780 r.p.m. 


■ Peripheral speed of armature 


1570 feet per 


2455 feet per 




minute 


minute 


Armature 






External diameter 


3 inches 


12 inches 


Internal diameter of laminations 


1 inch 


4j inches 




(shaft diameter) 




Gross length of core 


2 inches 


7| inches 


Length occupied by air ducts 





0.787 inches 


Type of slot 


overhung 


open (parallel sides) 


Number of slots 


24 


43 


Ratio average slot width to slot 






J pitch 


0.635 


0.442 


Type of tooth 


parallel sides 


tapered 


f Flux density teeth necks 


102700 lines 


117300 lines 


1 


per square inch 


per square inch 


Flux density armature core 


53000 lines 


67700 fines 


1 


per square inch 


per square inch 


Iron losses 


2 per cent input 


2 per cent input 


Inductors per slot 


18 


8 


Current density 


3000 amperes 


2770 amperes 




per square inch 


per square inch 


* Ampere inductors per inch of 






periphery 


185 


456 


Armature slot space factor 


0.34 


0.34 


Ratio armature IR drop to line 






voltage 


6.6 per cent 


3 per cent 


Armature PR loss 


6 per cent input 


3 per cent input 


Ratio iron loss to copper loss 


0.33 


0.66 


Ratio weight active iron to copper 


1.8 


3.75 


Field 






Ratio pole arc to pole pitch 


0.678 


0.75 


Radial depth of air gap 


0.030 inch 


0.150 inch 


Mean gap flux density 


30550 lines 


43200 lines 




per square inch 


per square inch 


Ratio ampere turns for gap to total 






ampere turns 


0.80 


0.545 


( Ratio field ampere turns to arma- 






ture ampere turns per pole 


1.50 


1.74 


Current density field coils 


2173 amperes 


1116 amperes 




per square inch 


per square inch 


Field PR loss 


6.6 per cent input 


1.7 per cent input 


Space factor shunt winding 


0.44 


0.5 


' Performance 






Efficiency full load 


75 per cent 


90.8 per cent 


Efficiency half load 


65 per cent 


87.2 per cent 


Efficiency quarter load 


56 per cent 


79.5 per cent 


Total weight per hp. output 


320 pounds 


70.6 pounds 


' Approximate cost materials per hp. 






, output 


$20.00 


$3.10 


Approximate selling price per hp. 






output 


$112.00 


$14.00 


♦ See Hobart'3 "Electric Motors," p. 153. 


9 





10 DESIGN OF SMALL IMOTORS 

Comparison between Small and Large B.C. Motor, In Table I 
a comparison is made between a small and a large motor which 
shows the influence of size on the current and flux densities in 
the various parts, the magnitude and distribution of losses, and 
certain significant ratios, dimensions, etc. As this table is not 
intended as a complete specification only such information is given 
as will illustrate the important characteristic differences between 
large and small motors. 

Rating. It will be noted that two poles are employed on 
the small motor as compared with four on the large motor. The 
choice of the number of poles involves a compromise between the 
simplicity secured with a small number of poles and the reduced 
size and weight obtained with a greater number of poles. 

The choice of voltage is not always wathin the scope of the 
designer, as frequently the motor must be run on a given source 
of power the voltage of which may have been previously deter- 
mined to suit other purposes. Li cases where the designer is able 
to choose the voltage, it should be higher with the larger motors. 
This will avoid the use of excessively fine wire on small motors 
or excessively large and expensive commutators on large motors 
which would be the case if the voltages w^ere the same. 

The speed of the small motor is 2000 r.p.m. as compared with 
780 r.p.m. for the large motor. The peripheral speeds, however, 
are 1570 and 2455 feet per minute respectively. The choice of 
speed is determined by mechanical requirements and by commuta- 
tion conditions. The centrifugal force per pound of material at 
the armature and commutator surfaces must be kept well within 
the strength of the materials. If the same factor of safety were 
used, the r.p.m. of a motor would be inversely proportional to the 
square root of the diameter. It is usual, however, to employ 
greater factors of safety in the larger machines. The usual range 
of commutator peripheral speed is 800 to 4000 feet per second. 
The higher values ^should be employed only with commutators of 
large diameter, as the brushes do not remain in contact as well 
with the greater angle of cur^'ature of small commutators. 

Armature. The dimensions and ratios listed in the table 
under Armature illustrate several features of design which are 
typical of small and large motors, respectively. The small arma- 



10 



DESIGN OF SMALL MOTORS 11 

ture laminations are assembled directly on the shaft, while in the 
large armature they are mounted on a spider. The small arma- 
ture has no air ducts, while in the large one the greater ratio of 
volume to radiating surface requires this additional ventilation. 
The overhung or nearly closed slot is necessary in the small motor 
in order to keep the reluctance of the air gap low, while in the 
large motor there is no difficulty in providing sufficient field 
ampere turns to permit the use of open slots with the consequent 
advantage of employing form- wound armature coils. 

The electrical design constants are also significant. In the 
small armature, the best results are obtained by providing as 
much winding space as possible. It is seen that the ratio of the 
average slot width to the slot pitch is 0.635 as compared with 
0.442 in the large armature, This results in a lesser amount of 
iron compared to the amount of copper in the small motor. In 
addition to using relatively more copper in the small armature, 
the current density is higher. This results in an armature PR 
loss of 6 per cent as compared with 3 per cent for the large 
motor. 

The flux densities in the teeth and armature of the small 
motor are on the other hand lower than in the large motor. This 
is due primarily to the necessity for keeping down the number of 
field ampere turns required. 

Field. The most striking feature of the field design is that 
in the small motor the field PR loss is 6.6 per cent of the motor 
input as compared with 1.7 per cent for the large motor. It is 
for this reason that in the design of the armature it w^as necessary 
to do everything possible to keep down the number of field ampere 
turns which would be required. In spite of this, however, the 
field losses in the small motor are very high. It w^ill be noted 
that most of the ampere turns are required for the air gap. It is 
necessary^ to make the gap large enough to control the distorting 
effect of the armature reaction. The ratio of field to armature 
ampere turns should be at least 1.25 in small motors, while in the 
case of large motors in which a considerable m.m.f. is required for 
the iron parts of the magnetic circuit it is customary to make the 
ratio of ampere turns required for only the air gap and teeth at 
least equal to the armature ampere turns per pole. 



11 



12 DESIGN OF SMALL MOTORS 

Performance. The efficiency at full load is 75 per cent for 
the J-hp. motor as compared with 90.8 per cent for the 25-hp. 
motor, while at partial loads the difference is even greater. These 
differences are due to the use of high£r current densities in both 
armature and field windings as well as to the fact that the amount 
of material per horsepower output is 320 pounds for the small motor 
as compared with 70,6 pounds for the large motor. For the latter 



Fig. 7. Stator of Single-Phase Induction Motor 
Courtesy of HoUzer-Cahot Company, Boston, Massachusetts 

reason the selling price per horsepower is $112 for the small motor 
and $14 for the large motor. 

Induction Motors.* Polyphase induction motors are not used 
to any extent in sizes smaller than several horsepower. This 
treatment will therefore deal only with single-phase motors. The 
principle of the single-phase induction motor when running at 
full speed is the same as that of the polyphase motor; namely, the 
production of a rotating magnetic field which induces current in a 
secondary member usually equipped with a squirrel-cage winding. 

* Sec "Alternating-Current Macliiuery," Part VI. 



12 



DESIGN OF SMALL MOTORS 



13 



Torque is thus developed by the reaption between the secondary 
conductors and the rotating field. The single-phase motor requires 
special arrangements for starting. The methods most commonly 
in use are hand starting, the split-phase method, and repulsion 
motor starting. 

Hand Starting, Hand starting is accomplished by merely 
giving the belt a pull and closing the circuit. It is necessary to 
attain a certain critical speed before the motor will "pick up." 
This critical speed is higher the heavier the load. Hand starting 




1^ 


i 






ik^ 


J 




Fig. 8. Parts of Western Electric Type SA Induction Motor with Clutch 

is therefore unsuitable when there is considerable load at starting 
or when the heavy starting current is objectionable. 

Split-Phase Starting. Split-phase starting is accomplished by 
adding a second winding in the primary slots which is displaced 
in space (usually one-half a pole pitch) with respect to the main, 
or running, winding. Fig. 7 shows a wound stator of a six-pole 
motor made by the Holtzer-Cabot Company. The starting wind- 
ing is generally made of relatively fine wire having a high resist- 
ance, so that the current in it is substantially in phase with the 
line voltage. The current in the main winding at starting lags 
behind the voltage by a large angle owing to its lower resistance 
and greater reactance. These currents are therefore displaced in 



13 



14 



DESIGN OF SMALL MOTORS 



phase (in large motors nearly 80 degrees and in small motors 
between 45 and 70 degrees) and thus produce torque as in a poly- 
phase motor. When two-thirds full speed, or thereabouts, has 
been reached, the starting winding is usually cut out by a centrif- 
ugally operated switch. In Figs. 8 and 9 is shown the Western 
Electric type SA induction motor, which operates on this principle. 
Fig. 10 shows the Robbins & Myers induction motor with the 
centrifugal switch at the right end of the armature. 

The initial starting current in this type of motor is from four 
to six times the full-load running value. Fig. 11 shows a typical 




P r P 




3=21 L 



# 



k 




1 


Jl^lM 


m 




fnt^ 


S 


-D- 


wf 


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w 




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fe J Z w 
Y 


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T < 


? 

b 


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ii •! 



R^ 



Fig. 9. Western Electric Type SA Induction Motor without Clutch 

performance curve of a |-hp., split-phase, 60-cycle induction motor. 
Where split-phase motors have to start under considerable load, 
a clutch type is employed, as shown in Fig. 8. This allows the 
motor to accelerate before picking up the load. 

Rejpulsion Induction Motor. The heavy starting current 
taken by the split-phase induction motor is frequently objection- 
able when there are lights which are dimmed by the momentary 
drop in line voltage. In fact, the Commonwealth Edison Com- 
pany requires that all such motors over 1 hp. must be operated 
on 220- volt instead of 110-volt circuits in order to insure good 



14 



[ 



DESIGN OF SMALL MOTORS 



15 



Itage regulation of the system. The repulsion induction motor 
gives high starting torque with low starting current and at the 
same time has all the good "full-speed" qualities of the split- 
phase type. The repulsion type of motor is permitted on 110- volt 
circuits up to 3 hp. In Fig. 12 is shown a repulsion induc- 
Bdh motor* made by the Wagner Company. Fig. 13 gives the 
performance curves of a J-hp. motor of this type. The starting 
current is only 3| times normal with full voltage applied, while 
the starting torque is 4^ times normal. At 1450 r.p.m. the motor 







Fig. 10. Parts of Robbing & Myers Induction Motor, Frame 18, One-Fourth Horsepower, 
1750 R.P.M. , 60 Cycles, Single Phase 

becomes an ordinary induction motor and accelerates up to 1800 
r.p.m. synchronous speed. 

The principle of the repulsion induction motor is illustrated 
by Fig. 14, which shows a conventional field structure excited by 
alternating current. The rotor is a d.c. armature, which is shown 
as the "Gramme" type for simplicity but in practice is of the 
usual drum type. The alternating flux enters the armature at A 
and divides equally between the upper and the lower halves of 
the armature, leaving at B. Considering each half by itself, it is 
evident that an a.c. voltage will be generated within it as in a 
transformer. Since the two halves of the armature are similar, 
equal voltages will be generated within them; and by tracing out 

* Sec "Alternating-Current Machinery," Part I. 



15 



16 



DESIGN OF SMALL MOTORS 



the direction of the turns, it will be seen that at any instant the 
direction of the voltages is the same, so that when these halves 



20r 2000 



STAQTm CURRENT 
WJW STARTING WINDING 



15 



10 



1500 



-Q: lOOO 



500 




0.2 



WITH 5TARJIN6 WmiN6. 

I I 1^-^ I I I I I 



0.4 0.6 

TORQUE- LB. FT 



0.8 



1.0 



Fig. 11. Characteristic Curves of a One-Eighth Horsepower 60-Cycle, 4-Pole, 110-Volt, 

Self-Starting Induction Motor 

Courtesy of "Standard Handbook for Electrical Engineers," Jfth ed., page 535 

are joined together, no current flows in the windings. If a volt- 
meter were applied to the commutator across a horizontal diam- 
eter, it would show maximum 
voltage, but if applied across a 
vertical diameter it would show 
no voltage, while at an angle of 
45 degrees it would show an 
intermediate voltage. Suppose 
now that two brushes are located 
in the last-named position and 
short-circuited. A current will 
flow through the windings of 
suflficient value to consume the 
former voltage as impedance drop 
in the windings. At a certain instant of time the direction of the 
current in the outer conductors (the inner conductors are inactive 
since they do not cross the field flux) will be as shown in Fig. 15. 
Current flowing in a direction up through the plane of the paper 
is indicated by a dot, while that flowing downward into the paper 




Fig. 12. Single-Phase Motor 

Courtesy of Wagner Electric Manufacturing 

Company, St. Louis, Missouri 



16 



DESIGN OF SMALL MOTORS 



17 



70 






7 X 7BA -JH.P- 60^0-4 POLE- 110/220 VOLTS. 



m 



S5 20 

I 



40 



30[- 50 
20 



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, fr^-h^^^^^w . Ti 




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2(?(? 400 bOO 800 1000 IZOO 1400 1600 1800 

Fig. 13. Performance Curve of a Wagner Type BA Motor 




Fig. 14. Conventional Field Structure of the Repulsion Induction Motor Excited by 
Alternating Current 



17 



18 DESIGN OF SMALL MOTORS 

is indicated by a cross. Since all the conductors are carrying 
current in a magnetic field (except Nos. 2 and 6), each conductor 
will exert a force tending to turn the armature in the direction 
indicated by the small arrows. Nos. 5, 4? ^j and 8 tend to pro- 
duce clockwise rotation, while Nos. 1 and 5 exert forces in a 
counterclockwise direction. Nos. 2 and 6 exert no force, since 
they are out of the magnetic field. This leaves a balance in favor 
of clockwise rotation. If the brushes had been located across a 
horizontal axis, there would have been no resultant force, as there 
would be as many conductors exerting force in one direction as in 
the other. If the brushes had been located across a vertical 
diameter, there would have been no current, since there is no 
resultant voltage between these points. Therefore an oblique posi- 
tion is chosen. In the above discussion the armature has been 

considered stationary at a cer- 
^ -^^ tain instant of time during the 

a.c. cycle, but it is obvious that 
during the next half-cycle both 
the flux and the induced current 
will be reversed, and therefore the 
Fi-. 1.-). Direction of Currents in Rotor at torquc will bc iu the samc direc- 
tion as before. Thus this motor 
will iiccelerate and run as a repulsion motor. It would not be desir- 
able, however, to run the motor continuously under these conditions, 
as there would be commutation difficulties. It is therefore pref- 
erable to convert the motor into an induction motor after start- 
ing. This is done by a centrifugally operated short-circuiting ring, 
which connects all the commutator bars together and, in effect, 
converts the .armature into a squirrel-cage rotor. 

Co7)ipariso}i between Small and Large Motor. In Table II a 
comparison is made between a small and a large induction motor 
which shows the influence of size on the important fundamental 
features of design. 

Rating. The comments regarding voltage, speed,- and 
number of poles which are given in the discussion of Table I 
apply equally well to the comparison of small and large a.c. 
motors. The frequency, like the voltage, is usually fixed, and 
its choice is not within the power of the designer. In general. 



18 




TABLE 11 
Comparison of Small and Large Induction Motor 





i Horsepower 


50 Horsepower* 


Rating 






Horsepower output 


\ 


50 


Number of poles 


4 


8 


Voltage 


110 


440 


Frequency 


60 cycles 


60 cycles 


Number of phases 


1 


3 


Speed (synchronous) 


1800 


900 


Stator 






External diameter 


6.6 inches 


25 inches 


Internal diameter 


3.8 inches 


19 inches 


Frame length 


2.36 inches 


6.375 inches 


Length occupied by ducts 


none 


1,375 inches 


Number of slots 


36 


96 


Type of slot 


Overhung 


Open with 
paraUel sides 


Slot area 


0.25 inchXO.80 


0.32 inch X 1.5 




inch 


inches 


Ratio slot width to pitch 


0.753 


0.5.16 


Conductors per slot 


34 


6 


Space factor 


0.344 


0.35 


Air gap 


0.013 inch 


0.030 inch 


Crest value gap flux density 


25000 lines 


36500 lines 




per square inch 


per square inch 


Crest value density teeth at neck 


89400 fines 


83000 lines 




per square inch 


per square inch 


Crest value density stator core 


42600 fines 


64000 fines 




per square inch 


per square inch 


Iron loss in per cent of input 


8.5 per cent 


3.82 per cent 


Ampere conductors per inch pe- 






riphery 


175 


600 


Current density stator conductors 


2500 amperes 


2230 amperes 




per square inch 


per square inch 


Stator PR loss per cent of input 


11 per cent 


2.88 per cent 


Ratio iron loss to PR loss 


0.7725 


1.326 


Rotor 






External diameter 


3.774 inches 


18.94 inches 


Internal diameter 


|-inch shaft 


15.5 inches 


Number of slots 


25 


79 


Type of slot 


Circular 


Overhung 
0.4 inch X0.45 inch 


Conductors per slot 


1 


1 


Slot space factor 


1 


1 


Current density conductors 


1800 


2760 


Rotor PR loss including end rings 


6 per cent input 


3.48 per cent input 


Performance 






Efficiency at full load 


64 per cent 


87.6 per cent 


Efficiency at half load 


49.3 per cent 


78.7 per cent 


Power factor full load 


0.623 


0.903 


Magnetizing current in per cent 






total current 


58.6 per cent 


33.9 per cent 


Slip in per cent synchronous speed 


8 per cent 


3.8 per cent 


Ratio starting current to running 






current 


6 


6.69 


Weight per horsepower 


. 164 pounds 


50 pounds 


Approximate selling price per horse- 






power 


$124.00 


$15.80 



* See Gray's "Electrical Machine Design." 



19 



20 DESIGN OF SMALL MOTORS 

it would be preferable to use a lower frequency with larger 
motors. 

The number of phases is also a characteristic of the power 
supply, but if polyphase, the question will arise as to whether it 
will be preferable to use one or all phases for the motor. If the 
motor is over 1 hp., it will usually be best to use all the phases; 
if smaller, the issue will be decided by whether or not it will be 
worth while to run the extra wire or wires required by a poly- 
phase motor in order to gain its better efficiency and starting torque. 

Stator. As in the case of the d.c. motor, it is necessary to 
employ overhung slots in the small induction motor in order to 
obtain a reasonable magnetizing current. The air gaps in induc- 
tion motors may be made as small as mechanical considerations 
permit since it is not necessary to avoid the distorting effect of 
armature reaction which occurs in a d.c. motor. 

The gap flux density in the i-hp. induction motor is chosen 
considerably lower than in the 50-hp. motor in order to com- 
pensate for the inherently lower power factor of a small motor. 
This results in relatively higher copper losses compared to iron 
losses, although both are greater in percentage of input than cor- 
responding losses in the larger motor. 

The single-phase motor is equipped with a starting winding 
which occupies space in the stator slots and thus makes the motor 
larger for a given output than if it were polyphase. 

Rotor. A notew^orthy feature of rotor design w^hich is brought 
out by Table II is the very high current density employed in the 
large motor. In general, a lower current density would be 
expected, as in the 'stator windings. In this case, however, a high 
rotor loss is intentionally employed in order to obtain greater 
starting torque and to reduce the starting current. Even with 
this greater current density, the percentage loss is less than in the 
small motor. 

Performance. The comments on the performance of the small 
and the large d.c. motors apply equally well to the performance 
of the a.c. motors as regards efficiency, weight, price per horse- 
power, etc. The difference in power factor is noteworthy and is 
due to the much greater percentage of the total current which 
is employed as magnetizing current in the small motor. This is 



20 



DESIGN OF SMALL MOTORS 21 

in spite of the fact that lower flux densities were employed as well 
as overhung stator slots. 

The difference in slip is due to the greater percentage loss in 
the small motor. It should be noted, however, that the designer 
had to use a lower current density in the rotor of the small motor 
in order to obtain a sufficiently small slip, while in the large 
motor, a high current density was employed (in order to obtain 
sufficient starting torque) without giving much slip. 

The starting currents are in about the same proportion, but, 
as previously mentioned, special attention w^as required to obtain 
a sufficiently low starting current in the larger motor. 

Universal Motors. Universal motors are motors which are 
designed for operation on either alternating or direct current of 
the same voltage. Their largest field of use is in household 
appliances, such as sewing machines, vacuum cleaners, massage 
vibrators, fans, toys, etc. Other miscellaneous uses are at soda 
fountains for mixing drinks, in barber shop devices, etc. In 
general, they form part of some appliance which requires a small 
amount of power and the manufacture and sale of which is 
facilitated by having one type of motor for both alternating and 
direct current. They are generally wound for use on 110- volt 
circuits and are always of the series type. In reality they are 
a.c. commutator motors except that their small size permits the 
omission of compensating windings, etc. In larger motors com- 
pensating windings are necessary in order to obtain good commu- 
tation. Unless they are employed, the alternating flux induces a 
voltage in the armature coils under commutation, even when the 
brushes are located in the neutral zone A- A, Fig. 6. This is due 
to the fact that the magnetomotive force (m.m.f.) produced by 
the current in the armature is along the line of the brushes and 
thus a certain amount of a.c. flux passes through the coils, which 
are temporarily short-circuited by the brushes. This flux induces 
a voltage which will cause a current dependent upon the imped- 
ance of the coils. In a large motor the resistance of the coils is 
low, thus requiring a compensating field winding in series with 
the armature and displaced 90 electrical degrees from the main 
field winding. This compensating winding opposes the m.m.f. of 
the armature and thus no flux and no current are generated within 



21 



22 DESIGN OF SMALL MOTORS 



I 



the coils under commutation. In very small motors, however, the 
resistance of the windings is sufficiently high to hmit this current 
to a permissible value in the armature coils under commutation, 
and it is then possible to obtain satisfactory commutation without 
compensating field windings. Another feature of the universal 
motor is that it will always produce more torque under heavy 
loads on d.c. than on a.c. This is due to the fact that when 
operating on d.c, the voltage drop in the series field coils is due 
merely to their resistance, while on a.c. the voltage drop is larger, 
owing to their impedance. 

ILLUSTRATIVE DESIGNS 

Standardized Frames. In order to make small motors at 
minimum cost, it is customary- for manufacturers to standardize 
on a ''line" of frames for motors up to a certain size, beyond 
which another line of frames of somewhat different structure is 
adopted, and so on. Numerous patterns, dies, jigs, tools, etcr, are 
made up for each frame size, and it is only in exceptional cases 
that the use of other than these standard frames is justified. 
Thus in practice the work of the designer is to pick out the most 
suitable standard frame for any individual case and meet the 
requirements by applying special windings. 

Each frame has a corresponding saturation curve showing the 
effective flux developed in that frame as a function of the field 
ampere turns. This is obviously a characteristic of the structure, 
no matter what the windings may be, and is therefore listed with 
the other constants of that frame. These constants include all 
mechanical dimensions, heat-radiating capacity, permissible speed 
range, etc. With this information, the designer may choose the 
windings to meet any given case within the limits suitable to that 
frame size. 

In working out the various illustrative designs it has been 
assumed that the designer possesses a list of frame sizes with their 
corresponding constants, and the work is taken up from this 
point. As a matter of information, however, the original design 
of one size of frame is worked out in the section Design of 
Type SDA Frames in order to show the procedure which is fol- 



22 



DESIGN OF SMALL MOTORS 23 

lowed on the rare occasions when a new line of frames is 
designed. The following notation is used throughout: 

e= internally generated voltage or back-e.m.f. 
AT = number of inductors per slot 
*=flux in lines 

(R =flux density in lines per square inch 
r.p.m. = revolutions per minute 
P= number of pairs of poles 

c= number of paths in parallel in the armature winding 
m.m.f . = magnetomotive force, or ampere turns 

DIRECT=CURRENT MOTOR DESIGN 
yVHORSEPOWER 110= VOLT SHUNT MOTOR 
Requirements. The motor which is to be designed is an 
example of a very small d.c. motor which might be used for 
driving a small jeweler's lathe. 

Output :y^3-hp. at 3000 r.p.m. No-load speed not over 4000 r.p.m 
Requirements | Input: 110 volts d.c. supply. Efficiency at full load not less than 
. 40 per cent. 

Choice of Frame. In Fig. 16 is given a table of frame sizes 
of the Western Electric type SDA motors, which are suitable 
for motors from -^^ to sV hp. The smallest, frame No. 300, 
is 2fl inches in diameter. If the designer has had previous 
experience with this frame, he will know that it is too small 
to satisfactorily meet the requirements given. If he has not had 
such experience, he may start with this frame and work out the 
design to a certain point, but he will soon find that the heating 
is too great, or meet with some other obstacle which indicates 
that the No. 300 frame is too small. Suppose that he has done 
this by the method hereinafter described and decided that a 
larger frame is necessary. His calculations will have given him a 
basis for estimating how much larger frame is necessary. Assume 

further that he has found that the No. 300 frame gives -— hp. 

150 

with satisfactory heating. Then it would be evident that either 

the No. 305 or the No. 307 frame should be chosen, since the 

No. 315 frame is obviously larger than necessary. Suppose that 

he decides to try the No. 305 frame, which, it will be shown, 

proves to be satisfactory. He will look up the information which 



23 



24 



DESIGN OF SMALL MOTORS 



jcn 



to 



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* * * * 



24 



DESIGN OF SMALL MOTORS 



25 



i on file regarding this frame. This is as follows: saturation curve 
IS shown in Fig. 17; various mechanical dimensions as used here- 
inafter; and general recommendations: full-load speed, not over 
tOOO r.p.m., and total watts losses, not over 20. 

Flux. As is shown in Fig. 17, in order to work frame No. 305 
to the best advantage a flux of 35,000 Unes should be employed. 
A smaller flux would not use the frame to its full capacity, while 
any greater flux would be too far beyond the knee of the satura- 
tion curve and thus would require a large field m.m.f., which 
would not .give, much further increase in flux and would cause 



bO 








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40 



80 IZO 160 200 240 280 520 560 
M. M. F. IN AMPERE - TURNS PER POLE 

Fig. 17. Saturation Curve of Western Electric Type SDA, Frame 305 Motor, at No Load 



needless power losses in the field coils. The exact value of the 
flux may be varied somewhat, but in general it should be just 
over the knee of the curve. The design will therefore be worked 
out on the basis that the effective flux is to be 35,000 lines at 
full load at 3000 r.p.m. In order to prevent the no-load speed 
rising above 4000 r.p.m., it is evident that a shunt winding must 
be used. The ampere turns per pole required to produce 35,000 
lines of flux are 128 at no load. At fufl load there will be a cer- 
tain amount of demagnetizing and distorting reaction of the 
armature. The exact amount of this will depend upon the posi- 
tion of the brushes, but it will be sufficient to add 25 per cent, 
making 160 ampere turns. 



25 



26 



DESIGN OF SMALL MOTORS 



Field Coils. It is now possible to compute the field coils. 
The loss in any coil of a given mean diameter is proportionate 
to the volume of copper and the square of the current density. 
Thus, comparing two coils producing equal ampere turns but one 
double the cross-section of the other, the coil of the larger section 
has one-half the power loss, since the reduction in current density 
more than compensates for the increased volume of copper. Thus 
it is desirable, in order to secure the minimum field loss, to 
employ as large a cross-section of coil as is feasible. Referring 
to the punching shown in Fig. 6, it might appear that the entire 

6 A 



^50 

^ /oh 



5 
01 



10 

9 

8 

7 

6 

5 

4 

3 

2 

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500 1000 1500 2000 2500 
^00 4^00 4500 5000 5500 6000 
AMPERES P£R 5a, IN, 

Fig. 18, Copper Losses 



5000 
6500 



3500 A 
7000 5 



space contained within the two elliptical recesses could be filled 
with wire. It must be remembered, however, that the field coils 
are to be form wound and slipped on in order to obtain low 
manufacturing cost. Evidently, then, it is necessary to allow 
enough space so that after one coil is in place, the other may be 
slipped on without too much trouble or danger of injury. It is 
therefore wise to permit each 'coil to occupy only that portion of the 
elliptical area which is behind the flare of the poles. This area 
will be approximately equivalent to 0.35X0.4, or 0.14 square inch. 
Allowing for a 0.035-inch tape around the coil, the area will be 



26 



DESIGN OF SMALL MOTORS 27 

reduced to 0.315X0.365, or 0.115 square inch. Suppose that 45 
per cent of this section will be the total cross-section of copper, 
the remaining 55 per cent being occupied by insulation. This 
assumption will be checked later on. Thus the copper coil section 
will be 0.45X0.115, or 0.0518 square inch. Since each coil is to 

develop 160 ampere turns, the current density will be , or 

0.0518 

3090 amperes per square inch. The mean length of turn of the 

coil will be twice the sum of the frame length, the pole core arc, 

and the allowance for corners, or 2(1 + 1.50+0.80), which equals 

6.6 inches. 

The volume of copper per coil will be 0.0518X6.6, or 0.342 

cubic inch. Fig. 18 gives the watts per cubic inch as a function 

of the current density for commercial copper at 50° C. From this 

curve the power loss in the two field coils will be 2X0.342 cubic 

inch X 7.20 watts per cubic inch, or 4.925 watts. Since the voltage 

4 925 
is to be 110, the field current will be ' , or 0.0448 ampere. 

Since there are 3090 amperes per square inch, the cross-section of 

0448 
the field wire will be ' , or 0.0000145 square inch. 

ooyu 

The number of turns can be obtained by dividing the total 
copper section per coil, 0.0518 square inch, by 0.0000145, and the 
resistance by multiplying the resistance per foot by the length of 
mean turn times the number of turns. By referring to a wire 
table, it is found that the 0.0000145 square inch section comes 
between No. 37 and No. 38. Using 0.0518 square inch cross- 
section of coil, the use of these two sizes may be compared by the 
following tabulation: 

Hot Total 

Cross- Resistance Turns Resistance Current on Ampere Field 

Section per Foot per Coil Two Coils 110 Volts Turns Watts 

No. 37 0.00001557 0.5848 3325 2137 0.0514 171.0 5.66 
No. 38 0.00001235 0.7374 4190 ^ 3400 0.0323 135.3 3.55 

The original assumption was 160 ampere turns, but this 
number cannot be secured by varying the number of turns of the 
field coils. No more turns can be added, as there is no more 
room, and if turns are taken off, the resistance of the field circuit 



28 



DESIGN OF SMALL MOTORS 



will drop in about the same proportion as the number of turns 
and. hence the ampere turns will remain constant. It is therefore 
necessary to use either 135.3 or 171 ampere turns. As it would 
not be good policy to weaken the field flux and thereby overload 
the armature, the 171 ampere turns will be employed. The 

effective ampere turns will be -— ^, or 136.8. Referring to Fig. 17, 

it is found that the corresponding effective flux is 36,000 lines 
instead of 35,000, which was originally chosen. Since No. 37 
enamel-covered wire is to be used for the field winding, the pre- 
ceding tabulation shows that there will be 3325 turns per coil. 
Since the size of wire has been determined, the assumption that 



SPACE FACTOR FOR ENAMELED WIRE 



















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li6H7LV WOUND 
(MINIMUM INSULATION) 



LOOSELY WOUND 
(MAXIMUM insulation) 



Fig. 



5 10 15 20 25 30 35 40 MILS D/AM£T£R 

19. Space Factor for Determining Sectional Area Occupied by Insulated Wire 



45 per cent of the space occupied by the wire will be copper 
section may be checked. Fig. 19 shows that the space factor is 
0.35, or 35 per cent, for enamel insulation. The difference is not 
sufficient to substantially alter the length of mean turn or prevent 
putting on the same number of turns, although the coil will be a 
little larger. It is well to check the last computation of field 
loss by Fig. 18. The current density in the field coil will be 

0.0514 
00001557 ' ^^ ^'^^^' ^^'^^^^ ^^^^^ ^'^^ ^^^^^ P^^ ^"^^^ ^^^^' "^^^ 
volume of copper will be 0.342 cubic inch per coil. The watts 
loss will be. 2X0.342X8.35, or 5.7, which checks reasonably close 
to the figure of 5.66 given in the preceding tabulation. 



28 



DESIGN OF SMALL MOTORS 29 

- Armature Winding. Suppose that the armature IR drop will 
be 25 volts. This assumption will be revised later. There will be 
about 2 volts drop in the brushes, so that the effective, or back, 
voltage of the armature will be 110-25-2, or 83 volts. The effec- 
tive flux has been settled as 36,000 lines and the r.p.m. is given 
as 3000. The voltage equation is 

60 c - 

■from which can be obtained 

eXQOXcXW 



N'- 



$X r.p.m. X2P 



which equals 



.J 83X60X2XW ...„ . , ^ 
^ = 36000X3000X2=^^'^ ^"^^^^^^^ 



Since there ave to be 11 slots, there will be 420 inductors per slot. 

An ordinary simplex drum winding with a 5/11 pitch will be used, 

5/11 pitch meaning that one coil encloses 5 teeth out of the total 

number of 11. The slot cross-section is to be an irregular shape 

whose area will be approximately 0.05 square inch, as may be 

found by plotting on cross-section paper. The slot will be Hned 

with 0.015-inch treated paper, which will leave an area of 0.04 

square inch. Further allowance must be made for a slot wedge 

at the top, which will leave 0.035 square inch for the winding. 

Assume that 25 per cent of this, or 0.00875 square inch, will be 

'occupied by copper cross-section. It is not desirable to attempt to 

' crowd the slots in such a smalljBdtor on account of the difficulty 

^of winding. The cross-section of copper in one inductor will be 

,-^ — —— ^ , or 0.00002085 square inch. The nearest size of wire 

to this is No. 36, which is 0.00001964 square inch. Therefore a 
No. 36 wire insulated with enamel and with a silk covering o\'er 
the enamel will be used. This will prevent short-circuited turns, 
which would be serious in the armature. 

In Fig. 20 is shown the ratio of net cross-sectional area of 
round copper wire in any slot to the gross sectional area of the 
slot; this ratio is called the space factor. 



29 



30 



DESIGN OF SMALL MOTORS 



The length of mean inductor is the sum of the length of 
armature frame and the length of one end connection, or 1+L5, 
which equals 2.5 inches, or 0.2085 foot. The total length of wire 
will therefore be 0.2085X420X11, or 961.5 feet, the resistance of 
which length for No. 36 is 445.5 ohms. Since there are two 
paths in parallel of one-half this length, the resistance of the 
armature winding will be one-fourth of this, or 445.5 -^ 4, or 111.3 
ohms. 

Although the armature current cannot be determined until 
Aater, it will be well to check up roughly on the ratio of field 











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20 AO 60 80 100 120 140 160 100 200 220 MiLS. 




3025201816 1.4- 12 10 8 6 4- BSS. 



30 252018 16 14- 12 10 8 6 &.W.G. 

Fig. 20. Curves Showing Space Factors for Round Wires of Various Gages 

ampere turns per pole to armature ampere turns per pole by esti- 
mating the armature current. Assume that the efficiency of the 

motor will be 45 per cent. The input will then be —^0.45, 

75 . .. 
or 22.1 watts. The field watts have already been determined as 
5.7, leaving 16.4 for the armature circuit. This would give an 

16 4 
armature current of ~, or 0.149 ampere. With this current the 



number of armature ampere turns per pole will be 
420X11 1 



0.149 



X 



X 



2 inductors per turn 2 poles 



2 paths 
, or 86.1. The ratio of field ampere 



30 



DESIGN OF SMALL MOTORS 31 

171 
turns to armature ampere turns is therefore -— — , or 1.985. This 

86.1 

ratio should be over 1.25 in order to secure good commutation. 
Since it is large enough in this case, the design may be contin- 
ued with confidence that it will work out in a satisfactorily 
manner. Had this ratio been less than 1.25, it would have been 
desirable to reconsider the choice of flux and endeavor to employ 
more ampere turns on the field. This would give more flux and 
require fewer turns on the armature, both of which would tend to 
increase this ratio. If the ratio could not then be raised to 1.25, 
it would be necessary to employ a larger motor frame or increase 
the speed. 

Losses. The iron losses are divided into hysteresis and eddy- 
current losses, which are usually computed by empirical formulas. 
Fig. 21 gives the watts per pound iron losses for various flux 
densities and frequencies. Referring to Fig. 6, it is evident that 
the flux density will be very low throughout the armature except 
in the necks of the teeth. It is a good plan to first compute the 
losses in the teeth necks and then decide whether it is worth 
while to compute the core losses. The pole arc will be 135 
degrees. Therefore the average number of teeth under one pole 

will be irX-rT-, or 4.125 teeth. The flux density in the teeth 

2 180 

necks will be the effective flux divided by the section of 4.125 

teeth, which equals T-ji^r — -^y or 87200 lines per square inch. 

The volume of iron in the active teeth necks will be 2X4.125 
X0.1X0.25 XI, or 0.209 cubic inch, the weight of which will be 
0.0581 pound since the density of iron is 0.278 pound per cubic 
inch. At 3000 r.p.m. the frequency in the armature will be 
3000-^60, or 50 cycles, as the motor is to be a two-pole machine. 
Since, according to Fig. 21, the iron loss is 8.5 watts per pound, 
the loss in the teeth necks will be 8.5X0.0581, or 0.494 watt. 
Since the iron loss in the core will be still less, it would clearly be 
unprofitable to compute it, but in the case of a larger motor it 
would not be negligible. There will be some stray power loss 
which cannot be readily computed, and therefore 0.8 watt will be 
allowed for all iron and stray power loss. 



31 



32 



DESIGN OF SMALL MOTORS 






The brush friction loss is computed as follows: The brush 

5 5 
area will be — X — , or 0.0244 square inch. Although this 

oZ oZ 

brush is capable of carrying a larger current than will be used, 
it would be unwise for mechanical reasons to use a smaller one. 
The brush pressure will be 4 pounds per square inch. In larger 
motors this may be reduced to 2 pounds per square inch, but it 
would not be safe to attempt to use less than 4 pounds in this 



io UO 



FREQUENCY 

0^^15 20 25 30 40 50 60 



100 
80 



^ 60 



40 

20 





80 



100 





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0R6ENERATOJ?5 




















































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5 JO J5 20 25 50 35 40 45 50 



JO 15 20 25 30 35 40 45 
WATTS PER POUND 

Fig. 21. Iron Losses for Direct-Current Motors or Generators 



case. The coefficient of friction of the brush will be about 0.35. 

The surface speed of the commutator will be X , or 

60 12 

13.2 per second. Hence the power loss in both brushes will be 

2X0.0244X4X0.35X13.2, or 0.901 foot pound per second, which 

equals -J-— hp., and -i— -X74C), or 1.223 watts. 
550 ooO 

From the formula* 0.8 cU (—;) watts, may be calculated 

the bearing loss when d is bearing diameter in inches, / the bear- 
ing length in inches, and v the rubbing velocity in feet per minute. 
In this case the bearing diameter will be 0.22 inch, and the length 

O.G inch. The rubbing velocity will be -^— -X3000, or 173 



* Sec "Standard Handbook for Electrical Engineers," 4th ed., Sec. 8. H 140. 



32 



DESIGN OF SMALL MOTORS 33 

feet per minute. Thus the loss in two bearings will be 2X0.8X 
0.22X0.6X1.73% or 0.482 watt. This formula, however, is for 
a bearing without side pull due to the belt, and in a very small 
motor the belt pull is usually much greater than the weight of 
the armature. To allow for this extra load and the fact that with a 
wick bearing the lubrication is not as perfect as with an oil- 
well bearing, 1 watt will be added to the result just obtained, 
making the bearing loss 1.482. 

The windage loss will be practically negligible, but 0.25 
watt will be allowed for the sake of completeness. 

The next step is to determine the armature current. The 
output of the- motor is to be yV hp., which equals 746-^75, 
or 9.95 watts. The armature current will be sufficient to supply 
this output and the various losses in the armature. 

Thus the armature input (less the PR loss) will be as follows: 

Useful power output 9.950 watts 

Brush friction 1.223 watts 

Bearing loss 1.482 watts 

Windage 0.250 watt 

Iron and stray power loss 0.800 watt 

Total 13.705 watts 

The back-e.m.f. was assumed to be 83 volts. On this basis the 
armature current wil], be 13.705^83, or 0.165 ampere. Since the 
resistance of the armature winding was computed as 111.3 ohms, 
the IR drop in the armature will be 0.165X111.3, or 18.8 volts, 
showing that the first estimate of 25 volts was too high. Assume 
that the drop will be 19 volts, which means that the back-e.m.f. 
will be 89 volts. On recomputing the number of armature induc- 
tors the following result is • obtained : 

,. 89X60X2X10^ .^.,,. , ^ 
/ = 3600()X3000 ->C2^'''^'"^^^^^^^^ 

The number of inductors per slot will therefore be 4950-i-ll, or 
450, an increase of 7.2 per cent over the first value of 420. There 
is enough room in the slots to allow the use of the same size of 
wire. No. 36. As the armature resistance will be increased in 
slightly greater proportion^ than the number of inductors per slot, 
owing to the increased length of turn, 1 per cent will be added 



33 



34 DESIGN OF SMALL MOTORS 

for this. The new armature resistance will be 11L3X 1.082, or 
120.3 ohms. 

The new armature current will be 13.705 -^ 89, or 0.154 
ampere, and the IR drop 0.154X120.3, or 18.53 volts, which 
is sufficiently close to the assumed value of 19 volts. Using the 
corrected values, the armature ampere turns per pole will be 

0154^450X11 ^j_^ or 95.3. The ratio of field ampere turns to 
2 2 Z 

171 
armature ampere turns will be -— — , or 1.8, which is satisfactory. 

95.3 

The PR watt loss in the armature will be 0.1542x120.3, or 2.85 

watts, and the current density in the armature inductors 

— ^ , or 3925 amperes per square inch. Referring to 

Fig. 18, it is found that this density gives 11.45 w^atts per cubic 
inch. Since the length of mean inductor is 2.5 inches, the number 
of inductors 4950, and the wire cross-section 0.00001964 square 
inch, the volume of armature copper will be 2.5X4950X0.00001964, 
or 0.243 cubic inch. The PR loss will therefore be 0.243X11.45, 
or 2.78 watts, which checks the value of 2.85 by the other method 
within sufficiently close limits. The brush drop loss will be 2 
volts X0.154 ampere, or 0.308 watt. 

The various losses will be as follows: 

Field copper 5.660 watts 

Armature copper 2.850 watts 

Brush contact 0.308 watt 

Iron and stray pov.cr loss 0.800 watt 

Brush friction 1.223 watts 

Bearings 1.482 watts 

Windage 0.250 watt 

Total 12.573 watts 

The No. 305 frame is capable of dissipating 20 watts losses with- 
out exceeding 50° C. rise. The design is therefore safe as regards 
temperature rise. It is evident that this design would stand con- 
siderable overload. 

Summary. The output in mechanical power is to be ^ hp., 

or 9.95 watts. The efficiency will therefore be — ^^^ , or 

9.95+12.573 

44.2 per cent, which is satisfactory for a motor of this size. A 



34 



DESIGN OF SMALL MOTORS 35 

no-load speed requirement of 4000 r.p.m. or less must be met. 
It is evident that this design will do this, since the armature 
IR drop will be only 19 volts and, even neglecting the increase 
in flux, with a back-e.m.f. of 89 the speed will not be more than 

— ^X3000, or 3640 r.p.m. 

The shop information for this TF-hp. motor may be sum- 
marized as follows: 

Mechanical Data 

Frame No. 305, type SDA. Brushes grade E-1 **Le 

Carbone" ^X^ section with 1| ounces spring pressure. 

All other parts standard with No. 305 frame. 
Electrical Data 
Field Winding: 3325 turns per coil of No. 37 enamel-covered wire, 

standard tape wrapping. Coils connected in series. 

Field circuit in shunt with armature. 
Armature Winding: 450 inductors per slot of No. 36 enamel- and silk-eovered - 

wire. Simplex, singly re-entrant, 5/11 pitch. Standard 

slot Hning. 

This is all the information necessary if a standard line of 
frames is employed. 

Comments. The most striking feature of this design is the 
low value of the armature iron losses compared to the copper 
losses. It is worth while to inquire whether the efficiency of the 
motor could not be improved by some change which would reduce 
the copper losses and increase the iron losses, according to the 
general principle which applies to large machines that these 
losses should be nearly equal. The iron losses may be increased 
for a given armature, by increasing either the flux density or the 
frequency. It would not be profitable to increase the flux 
density in this case, since that would require increased field ampere 
turns and hence increased field losses. These losses are already the 
largest single item. The flux density in the teeth might be 
increased by employing narrower necks, but that would make 
them dangerously weak mechanically without giving very much 
more slot section. The speed could be increased and the fre- 
quency thus raised, obtaining a proportionate increase in the 
power output of the motor without increasing the copper losses. 
The objection to this procedure would be that at very high speeds 
the life of the commutator is short. Thus, for a motor to pro- 



35 



I 



36 DESIGN OF SMALL MOTORS 

duce a given horsepower, it is cheaper to buy a larger-frame 
slow-speed motor than to use a small-frame high-speed design 
which would wear out more quickly. 

yV=HORSEPOWER 12=V0LT SHUNT MOTOR 
Requirements. The purpose of this design is to illustrate 
the changes necessary in the preceding design when the supply 
voltage is only 12 volts instead of 110 volts. By comparing the 
following design with the preceding one and using the same pro- 
cedure the student will be able to compute windings for any 
voltage from 6 volts up to 110 volts. It would not be practical 
to attempt winding this very small motor for any voltage higher 
than 110, because too fine wire would have to be used. 

r Output: -^ hp. at 3000 r.p.m. No-load speed not over 4000r.p.m. 
Requirements \ Input: 12 volts d.c. supply. Efficiency at full load not less than 
^ 40 per cent. 

Field Coils, As the frame size for a motor of a given power 
output is practically independent of the voltage, the No. 305 
frame will be used. Owing to the lower voltage, the space factor 
of the field coils will be better, since a larger size of wire will 
be used. Fig. 19. A somewhat greater flux may therefore be 
employed in this than in the 110- volt design. Choosing $ as 38,000 
instead of 35,000, at no load, 160 ampere turns per pole, Fig. 17, 
will be needed, and at full load 160X1.25, or 200 ampere turns 
per pole. Referring to the previous calculations, it is found that 
the space available for the field coil section will be 0.115 square 
inch, of which the copper section will be approximately 0.0518 
square inch; that the length of mean turn will be 6.6 inches; 
and that the volume of copper will be 0.342 cubic inch. The 

current density in the field coil will be - — - — , or 3860 amperes 

0.0518 

per square inch, which from Fig. 18 involves a loss of 11.25 watts 

per cubic inch. The field watts for the two coils will be 2X0.342X 

11.25, or 7.69 watts. As a shunt winding is to be used, the field 

7 69 
current will be -j— , or 0.641 ampere. The section of wire will 

be 0.641 -j- 3860, or 0.000166 square inch. Referring to a wire 
table. No. 27 is 0.0001583 square inch and No. 26 is 0.0001996 



36 



DESIGN OF SMALL MOTORS 37 

square inch. If No. 26 is used, a lower-resistance winding is 
obtained than with No. 27, and hence greater losses. As the 
losses are already Jiigh No. 27 will be used. 

The number of turns will be ^ ^ ' , ^^^ , or 328.5, which will 

0.0001o83 

be considered 328 turns per coil. The length of wire per coil will 

be 328X^, or 180.5 feet. The resistance per foot for No. 27 

is 0.0575 ohm. The resistance for two coils will be 2 X 180.5 X 
0.0575, or 20.75 ohms. Since the voltage is to be 12, the current 
will be 12 -^ 20.75, or 0.578 ampere. The watts loss will be 
12X0.578, or 6.93 watts. The ampere turns per coil will be 
0.578X328, or 189.5. The equivalent no-load ampere turns will 
be 189.5-^1.25, or 151.8. From Fig. 17 the corresponding flux 
is 37,500 instead of the origiqal assumption of 38,000. A flux 
of 37,500 will be satisfactory, however. 

Armature Winding. Metallized carbon brushes will be used 
and will have a contact drop of 0.8 volt. If the IR drop in the 
winding is estimated as 2 volts, the back-e.m.f. will be 12 — 0.8 
— 2.0, or 9.2 volts. From the voltage equation 



can be obtained 



which becomes 



,=iv$x^^x— xio-« 

60 c 



^ eXQOXcXW 



*Xr.p.m.X2P 



NJ^^^^^^m inductor. 
37500X3000X2 

Since there are to be 11 slots, there will be 491-^ll, or 44.7 induc- 
tors per slot. As a simplex winding with two coils per slot will be 
used, an even number of inductors will be needed and therefore 
44 inductors per slot will be employed. 

From the previous design it is found that the windmg space 
per slot will be 0.035 square inch. Assume that 30 per cent of 
this will be copper section instead of 25 per cent, since a larger 
wire will be used than with the 110- volt design. The inductor 

section will thus be — — — -^ , or 0.000239 square inch. No. 25 



37 



38 DESIGN OF SMALL MOTORS 

has a section of 0.0002517 square inch. This is only slightly 
larger, and therefore it can be used and, if necessary, the wedges 
in the top of the slots can be omitted, as they are not essential 
with this larger wire. The length of mean inductor will be 2.5 
inches, or 0.2085 foot. The total length of wire will be 0.2085 X 
44X11, or 101 feet. Since the resistance of 1 foot of No. 25 is 

0.0362 ohm, and there will be two paths in parallel — feet long, 

the resistance of the armature will be 10lX0.0362-^4, or 0.915 
ohm. 

Losses. The various losses will now be computed in order 
to obtain the armature current. The iron losses will be about 
the same as in the 110-volt design since the final flux is 37,500 
compared with 36,000. As these losses are so small, the iron and 
stray power losses will be estimated as 1 watt. 

The brush friction loss depends upon the size of brush, and 

that, in turn, depends upon the current, which is not yet computed. 

It can be estimated sufficiently closely, however, to select the 

brush section. It is known that the over-all efficiency of the 

motor will be about 45 per cent, which will give an input of 

746 

— — -^0.45, or 22.1 watts, which means 22.1-^-12, or 1.845 amperes. 

Subtracting the field current of 0.578 ampere, the approximate 
armature current of 1.267 amperes is obtained. The metallized 
carbon brush will easily carry 75 amperes per square inch, giv- 
ing a brush section of 1.267 -^ 75, or 0.0169 square inch. For 
mechanical reasons it is not advisable to use a brush smaller 
than y^XA, or 0.0244 square inch, which is more than the 
necessary area. As this is the same size of brush as in the pre- 
ceding 110-volt design, there will be about the same friction loss 
— 1.22 watts. The same bearing loss — 1.48 watts — and the 
same allowance for windage — 0.25 watt — will be used. 

The armature input (less the PR loss) will be as follgws: 

Useful power output 9.95 watts 

Brush friction 1.22 watts 

Bearing loss 1.48 watts 

Windage 0.25 watt 

Iron loss 1.00 watt 

Total ". . 13.90 watts 



38 



I 



DESIGN OF SMALL MOTORS 39 



e back-e.m.f. was assumed to be 9.2 volts, which would give a 
current of 13.9 -J- 9.2, or L51 amperes. Since the armature resist- 
ance will be 0.915 ohm, the IR drop in the armature will be 
1.51X0.915, or 1.382 volts, instead of the estimate of 2 volts. 
The back-e.m.f. will therefore be greater, which will require a 
correspondingly greater number of inductors. But as the first 
estimate of 44 inductors per slot of No. 25 wire already exceeded 
the permissible copper section per slot, it will be necessary to use 
No. 26. Assume 46 inductors per slot of No. 26. According to 
the voltage equation, the back-e.m.f. will be 

46X11X37500X3000X2 ^ .„ ,^ 

e = — - — - — — — = 9 .49 volts 

60X2X10^ 

The current will be 13.9-^9.49, or 1.465 amperes; and the resist- 
ance of the armature will be 46X11 X0.2085X0.0456-r-4, or 1.225 
ohms. The IR drop will be 1.465X1.225, or 1.795 volts, and the 
PR loss will be 1.4652X1.225, or 2.62 watts. The armature 

ampere turns per pole will be — - — - — ^ — , or 92.5. The ratio of 

field ampere turns per pole to armature ampere turns per pole will 

be -zr^, or 2.05, which is satisfactory since it is greater than 1.25. 

Allowing 0.8 volt brush contact drop, the impressed voltage should 
be 1.795-1-0.8+9.49, or 12.085 volts. Since the impressed voltage 
is to be 12, the speed will be a trifle less, but it will not be worth 
while to attempt to refine the calculations further. 
The various losses will be as follows: 

Field copper 6.93 watts 

Armature copper 2.62 watts 

Brush contact 0.8X1.465 1.17 watts 

Iron and stray power loss 1.00 watt 

Brush friction 1.22 watts 

Bearings 1.48 watts 

Windage 0.25 watt 

Total 14.67 watts 

Summary. Since the useful power output is to be 9.95 watts, 

9 95 
the efficiency will be ^ ^^ \ , ^^ , or 40.4 per cent, which just meets 
9.95 + 14.67 

the efficiency requirement. Since the losses will be less than 20 

watts and well distributed, the heating will not be excessive. The 



39 



40 DESIGN OF SMALL MOTORS 

only other requirement is a speed of less than 4000 at no load. 
Since the armature IR drop will be so small, this will be easily met. 
The shop information for the 12-volt rVbp. motor may be 
stated as follows: 

Mechanical Data 

Frame No. 305, type SDA. Brushes grade KK-4 "Le 

Carbone" ^X^ section with 1^ ounces spring pressure. 
All other parts standard with No. 305 frame. 
Electrical Data 
Field Winding: 328 turns per coil of No. 26 single cotton-covered wire, 

standard tape wrapping. Coils connected in series. 
Field circuit' in shunt with armature. 
Armature Winding: 46 inductors per slot of No. 26 double cotton-covered 
wire. Simplex, singly re-entrant, 5/11 pitch. Standard 
slot lining. 
Example: Work out and tabulate corresponding shop information on a 
24-volt, yV^P-' ^^^^ r.p.m. shunt motor. 

yio=HORSEPOWER SERIES MOTOR 
Requirements. This yio^ hp. motor might be used for driving 

a fan. 

Output: Y^ hp. at 3000 r.p.m. Must not overheat when apphed 
load is such as to reduce speed to 1500 r.p.m. No-load 
Requirements I speed must not exceed 8000 r.p.m. 

Input: 110 volts d.c. supply. Efficiency at full load not less 
than 35 per cent. 

Owing to the half-speed requirement against overheating, the 
frame size will have to be large enough to dissipate the additional 
heat developed under this condition. The No. 305 frame, type 
SDA, Fig. 16, is a good size to try, since it proved capable of 
yV hp. at 3000 r.p.m. in the previous shunt-motor design. At half 
speed the friction losses and the iron losses will be reduced by 
one-half, so that the copper losses will be the determining factor. 
These losses should be kept low under normal speed in order to 
allow for the increase at half speed. The field circuit should also 
be worked below the knee of the saturation curve so that under 
half-speed conditions the increased current will cause an increase 
of flux which will prevent the back-e.m.f. from falling to a point 
where the current would be excessive. 

Preliminary Design. Referring to Fig. 17, a normal flux of 
32,000 lines, requiring 103 ampere turns, is chosen. Adding 25 
per cent to this for armature reaction, 1.25X103, or 128.8 full- 
load field ampere turns, are obtained. In order to secure. good 



40 



DESIGN OF SMALL MOTORS 41 

commutation, it is desirable to make the field ampere turns per 
pole over 1.25 times, say 1.5 times, the armature ampere turns 
per pole. Since in a series motor the armature current in a con- 
ductor is one-half the field current, the ratio of field turns per 
pole to armature turns per pole will be 1.5X0.5, or 0.75. 

A trial value of current must now be assumed and it will be 
based on an eflSciency estimate of 40 per cent. The input of the 

746 
yio-hp. motor will thus be t7;7:"^0.40, or 18.65 watts. Since the 

voltage is to be 110, the current wdll be -j^jr, or 0.1695 ampere. 

128 8 
The field turns will be -.^^^ > or 761 turns per pole. Using the 

u.ibyo 

ratio determined above, the armature turns per pole will be — — -, 

0.75 

or 1015. Since there are two inductors per turn and two poles, 
the total number of inductors will be 1015X2X2, or 4060 induc- 
tors. The back-e.m.f. of the motor will therefore be 

60 c 
which equals 

e = 4060x32000x50XlXl0-8 = 65 volts 

The back-e.m.f. will therefore be — — , or 59 per cent of the 

impressed voltage. This is a reasonable value for the back-e.m.f. 
in such a small series motor. If it had come out much less, a 
greater flux would have been employed, and vice versa. This 
leaves 110 — 65, or 45 volts, to be consumed as IR drop in the 
armature and the field. The resistance of the motor will thus be 

45 
01695' "' ^^^-^ '^'''- 

Half=Speed Conditions. Before going further it will be well 
to make an estimate of the conditions at half speed, so that work 
may be saved in case it is necessary to modify the original 
assumptions. At half speed the back-e.m.f. will be less than 
normal but more than half normal, since the flux will be greater 
owing to the greater current. The saturation curve for frame 
No. 305, however, shows that the flux will not be very much 



41 



42 DESIGN OF SMALL MOTORS 

greater, as the knee of the curve will be passed. Assume that 
the flux will increase from 32,000 to 38,000 lines. The back-e.m.f. 
at half speed wall then be 

e = 4060 X 38000X25X1X10-8 = 38.5 volts 
The IR drop will be 110-38.5, or 71.5 volts, and the current 

will be -^^i^, or 0.2695 ampere. Since there are to be 761 field 
265.5 

turns per pole, the field ampere turns per pole will be 0.2695X761, 

205 
or 205, which corresponds to — — , or 164, at no load. Fig. 17 

shows that this value of ampere turns gives 38,500 lines, or slightly j 
more than the number assumed. This will give slightly greater 
back-e.m.f. and slightly less current, but it will not be worth while 
to recompute the design, as only preliminary data is desired now. 
The PR losses at half speed will be 0.2695^X265.5, or 19.3 watts. 
There will also be friction losses of a few watts, making the 
probable total losses at half speed about 24 watts. 

The data sheet on the No. 305 frame shows that it is rated 
to dissipate 20 watts with a 50° C. rise. Although this design will 
probably exceed this rise a little, the excess wdll not be large, 
and a little higher temperature rise then normal can be permitted, 
as it would be exceptional for the load to be maintained indefi- 
nitely at half speed. If the losses had been much too 
high, it would have been necessary to assume a full-load flux 
farther down on the knee of the saturation curve so as to give 
a greater percentage increase in flux with the increased current 
at half speed. This would give greater back-e.m.f. and less PR 
losses, but the motor would not develop as much power at nor- 
mal speed. 

No=Load Speed. The brush friction loss is of particular 
importance in this design, as it is this loss which will be used to 
limit the no-load speed to the required speed of 8000 r.p.m. The 
operation under no load is computed as follows: Assume a value 
of 95 volts back-e.m.f. since the current and the IR drop will 
be small. The flux may be determined from the voltage equation 

60 c 



42 



I DESIGN OF SMALL MOTORS 43 

and, on substituting values and solving for $, is found to be 

17,500 lines. According to Fig. 17 this corresponds to 42 ampere 

turns per pole. Adding 25 per cent to this for armature reaction, 

42X1.25, or 52 ampere turns, will be obtained. Since there are 

52 
to be 761 turns in each field coil, the current will be — — , or 

761 

0.0683 ampere. The IR drop in the motor will be 0.0683X265.5, 
or 18.15 volts, which will give a back-e.m.f. of 110 — 18, or 92 
volts. This is not far from the assumption of 96 volts. Pro- 
ceeding therefore on the basis of 0.0683 ampere at 95 volts back- 
e.m.f., the power developed by the armature will be 6.49 watts. 
Allow 1.49 watts as a minimum value for iron, bearing, and windage 
losses, and assume that 5 watts is to be dissipated by brush friction. 

33000 
The power will be 5X , or 221.3 foot pounds per minute. 

The commutator will be 1 inch in diameter, and hence the periph- 
eral velocity at 8000 r.p.m. will be -^X8000, or 2095 feet per 

221 3 
minute. The tangential friction force will therefore be ' , 

2095 

or 0.1054 pound. Assuming a coefficient of friction of 0.3, the 

radial force will be 0.1054-^0.3, or 0.351 pound, which makes 

0.1755 pound, or 2.73 ounces, per brush. If a brush pressure of 

4 pounds per square inch is employed, the brush area will be 

0.1755 -^ 4, or 0.04385 square inch. Make the brushes i^Xi, giving 

an area of 0.0469 square inch and a pressure of ' , or 3.74 

O.U'iUi/ 

pounds per square inch; and use a grade E-1 *'Le Carbone" 
brush. It is evident that the no-load requirement of less than 
8000 r.p.m. can be met. 

Armature Windings at Normal Speed. Returning to the 
design at normal speed, it may be continued with more confidence 
in the outcome. The armature windings will first be computed. 
There are to be 4060 inductors, which means 4060-^11, or 369 
inductors per slot. As there must be an even number, use 370 
inductors per slot and 4070 inductors in all. A simplex singly 
re-entrant winding will be used, cording the coils to a 5/11 pitch, 
as in the case of the yV-hp. shunt motor. As has already been 



4Z 



44 DESIGN OF SMALL MOTORS 

worked out in the design of the shunt motor, the total permis- 
sible copper section per slot will be 0.00875 square inch as obtained 
from the slot dimensions. The inductor section will therefore 

be ?:55?Z£, or 0.00002365 square inch. The wire nearest to this 
370 

size is No. 35, w^hich is 0.00002476 square inch. This is a trifle 

greater section, but as considerable slot room was allowed in the 

previous design, it k certain that with the better space factor of 

the larger size of wire it will be satisfactory. The length of mean 

inductor w^as found to be 2.5 inches, or 0.2085 foot. The total 

length of armature wire will therefore be 4070X0.2085, or 848 

feet. The resistance per foot of No. 35 is 0.3678 ohm (hot). 

The resistance of the armature will be one-fourth of the total 

wire since there are tw^o paths in parallel. Thus the armature 

resistance will be iX848X0.3678, or 77.9 ohms. 

Losses. The iron losses in the teeth necks are obtained from 

32000 
Fig. 2L The flux density in the teeth necks wall be ^ ,,^, lines, 

0.412D 

or 77600 lines per square inch. The volume of iron in the active 

teeth necks will be 2x0.4125X0.25X1, or 0.209 cubic inch, and 

the weight will be 0.209X0.278, or 0.0581 pound. From Fig. 21 

the iron loss per pound at 50 cycles and 77,600 lines per square 

inch is 6.5 watts per pound. The watts iron loss will thus be;j 

6.5X0.0581, or 0.3775 watt. Since the iron loss in the armature 

core will be still less, it will not be worth while to compute it. | 

The next step is to compute the brush friction loss at the 

normal speed of 3000 r.p.m. At this lower speed the coefficient 

of friction will be 0.35 as a maximum figure. The watts loss will 

be 2X0.1755X0.35X3000 X^X^^, 2.190 watts. 

The armature input less the PR loss at 3000 r.p.m. will ]&e 

as follows: 

Power output 7.460 watts 

Brush friction 2.190 watts 

Bearing loss* 1.482 watts 

Windage* 0.250 watt 

Iron loss and stray loss (estimated) 0.650 watt 

Tolal 12.032 watts 



* See the design of the ?'5-hp., 110-volt, shunt motor. 

44 



^DESIGN OF SMALL MOTORS 45 

Since the back-e.mi. has been found to be 65 volts, the current 

12 032 

input will be ^ — , or 0.1852 ampere. This is somewhat greater 

65 

than the original estimate of 0.1695 ampere. The resistance of 

the armature was previously computed as 77.9 ohms. The PR 

loss will therefore be 0.18522X77.9, or 2.67 watts. Since the 

section of No. 35 wire is 0.00002476 square inch, the current 

density in the armature will be — — nnnn9A7A ' ^^ ^^^^ amperes 

per square inch. From Fig. 18 the PR loss will be 10.5 watts 

per cubic inch, and since each of the 11 slots is to have a section 

of 0.00916 square inch, and the length of mean inductor is to be 

2.5 inches, the total PR loss will be 10.5X0.00916X11X2.5, which 

equals 2.64 watts. The armature IR drop will be 0.1852x77.9, 

or 14.44 volts. 

Field Coils. The field coils are to produce 128.8 ampere 

turns per pole. With the new value of current the turns will 

128 8 
be reduced from 761 to -.oro ? ^^ ^9^- '^^^^ ^^^^^ reduce the ratio 

0.1852 

of field ampere turns to armature turns a, trifle but not enough 

to be serious. From Fig. 6 the length of niean turn may be 

determined to be 6.6 inches. Thus the total length of wire in the 

field will be 2XYr-X696, or 766 feet. The voltage drop across the 

two field coils will be 110 volts — 65 (back-e.m.f.) — 14.44 (arma- 
ture IR)—2 (brushes), which equals 28.56. The resistance of 

28 56 
the field coils will therefore be .'ro ' ^^ 154.2 ohms. The resist- 

0.1852 

154.2 
ance per foot of wire will be ^' , or 0.2015 ohm. 

766 

Referring to a wire table, we find that this resistance comes 

between No. 32 and No. 33. Keeping the required 696 turns, 

field coils of the two sizes would compare as follows: 



Turns 

No. 32 696 

No. 33 696 

From the dimensions of the punching it will be found that either 



45 



Copper Section 


Total Section 


Resistance of 


per Coil 


of Taped Coil 


Two Coils 


0.03455 


0.1099 


140.3 


0.02740 


0.0905 


176.8 



46 DESIGN OF SMALL MOTORS 

of these coils will fit. The higher resistance of the No. 33 will 

reduce the total current and hence give less heating at half speed. 

Therefore use 696 turns of No. 33. The PR loss will be 6.08 

watts. As the IR drop will be 32.7 volts instead of the previous 

estimate of 28.56, the back-e.m.f. will be a trifle less and give a 

slightly lower speed than 3000 r.p.m. But as a series motor does not 

have an inherent exact speed, this will not be of any consequence 

and there is no need to revise the calculations on this account. 

Revised Half=Speed Data. It will, however, be advisable to check 

over the heating at half speed . Using the previous value of 38,500 lines 

e = 4070X38500X25XlXl0-« = 39.12 volts 

The IR drop will then be 110-2 (brushes) -39.12 (back-e.m.f.), 

which equals 68.88 volts. Since the armature resistance is to be 

77.9 ohms and the field resistance 176.8 ohms, the total resist- 
no oo 

ance will be 254.7 ohms, and the current will be r-Vr,, or 0.27 

254.7 

27x696 
ampere. This will produce ' — , or 150 ampere turns per 

l.Zo 

pole, which from Fig. 17 will give 37,500 lines. This flux value 
is sufficiently close to the 38,500 fines previously found; but the 
current will be slightly greater than 0.27 ampere. The PR losses 
in the motor at half speed will be 0.272x254.7, or 18.55 watts. 
Referring to the table of losses at normal speed, the friction losses 
due to brushes and bearings may be estimated as 2 watts and the 
other losses as 0.5 watt, making the total losses at half speed 
21 watts, which is only 1 watt in excess of the rating given the 
No. 305 frame for a 50° C. rise. The design may therefore be 
considered satisfactory, as half speed will not be maintained 
indefinitely. The maximum speed requirement of 8000 r.p.m. 
at no load will not be appreciably affected by any of the deci- 
sions since the design was first computed. 

The losses at normal speed will be as follows: 

Field copper 6.08 watts 

Armature copper 2.67 watts 

Brush contact 0.37 watt 

Iron and stray power loss 0.65 watt 

Brush friction 2.19 watts 

Bearings 1.48 watts 

Windage 0.25 watt 

Total 13.69 watts 



46 



1^ Summ£ 



DESIGN OF SMALL MOTORS 47 



Summary. The output will be rsir hp-, or 7.46 watts. The 

7 46 
efficiency will be ' , or 35.3 per cent, which just meets 

/ .40 ~r~ lo.oy 

the efficiency requirement. 

The shop information on the rJir-hp. motor may be sum- 
marized as follows: 

Mechanical Data 

Frame No. 305, type SDA. Brushes grade E-1 "Le 
Carbone" ^Xl section with 2| ounces spring pressure. 
All other parts standard with No. 305 frame. 

Electrical Data 
Field Winding: 696 turns per coil of No. 33 enamel-covered wire, stand- 

ard tape wrapping. Connect coils in series. Field cir- 
cuit in series with armature. 
Armature Winding: 370 inductors per slot of No. 35 enamel- and silk-covered 
wire. Simplex, singly re-entrant, 5/11 pitch. Standard 
slot lining. 

Example. Work out and tabulate shop information on a 12-volt y^o'hp- 
motor of similar requirements. 

DESIGN OF TYPE SDA FRAMES 

Magnetic Circuit. In the illustrative designs of the TF-hp. 
and the xio-hp. motors which used the No. 305 frame, it was 
assumed that the designer could obtain these frames from stock 
with information on the saturation curve, heating constants, etc. 
From this basis the designs were worked out. It will now be 
shown how these frames are designed. The No. 305 frame will be 
used to illustrate the design of the various sizes of type SDA 
listed in Fig. 16. 

In general, the field and armature punchings should be so 
proportioned as to give as much winding space as is possible 
within the outside diameter of the field punching. The design 
should also be such as to give an easy path for the flux so as to 
require the minimum number of ampere turns on the field coils. 
The magnetic circuit will be analyzed by computing the ampere 
turns required for each part, and then the effect of any change 
in the design will be studied. This will show that the present 
form is excellent when all the various points are considered. 

In Fig. 6 the path of the lines of flux is indicated by the 
dotted lines. Consider that the upper fleld coil supplies the 
ampere turns for driving the flux from the mid-points of the two 



47 



48 DESIGN OF SMALL MOTORS 

yokes on the line A A through the yokes to the field pole and 
across the upper air gap through the teeth and the armature to 
the middle of the armature core on the line AA. The lower 
field coil drives the flux through the lower half of the magnetic 
circuit in the same way. The two yokes form two similar paths 
in parallel between the poles. It is therefore sufficient to consider 
the upper right-hand quarter of the frame; and the ampere turns 
required to drive the flux from the mid-point of the yoke on the 
line A A to the mid-point of the armature core on the line A A 
will be the ampere turns which each field coil must develop. 
Assume a fiux density of 29,000 fines per square inch in the air 
gap, which multiplied by the area of the gap gives 40,000 lines 
gap flux. 

The following tabular matter shows the cross-section of each 
part of the circuit, its flux density, its length, and the ampere 
turns required to force the flux through that part. 









Flux Density 


Ampere 


Length 


Ampere 






Cross- 


in Lines 


Turns 


of Part 


Turns 


Part 


Flux in Lines 


Section 


per Sq. Inch 


per Inch 


(in.) 


per Pole 


Yoke 


42500^2 


0.25X0.95 


89600 


31.5 


1.1 


34.65 


Pole 


45000 


1.0X0.95 


47350 


5.0 


0.5 


2.50 


Air gap 


40000 


1.45X0.95 


29050 


9090.0 


0.0125 


113.70 


Teeth necks 


38500 


0.4125X0.95 


98300 


58.0 


0.2 


11.60 


Armature core 


38500 H- 2 
Total . 


0.25X0.95 


81100 


20.0 


0.6 


12.00 




. 174.45 



The flux is different in the different parts on account of the 
leakage flux. The amount of leakage is estimated. The cross- 
section is obtained by measurement of the various parts. The 
length of the frame used in computing the section was taken as 
0.95 inch instead of 1 inch in order to allow for the minute spaces 
between laminations. The flux densities are obtained by dividing 
the flux by the corresponding section. The ampere turns per 
inch are obtained from Fig. 22 with the exception of the air gap, 
for which the formula is ampere turns per inch equal 0.313 times 
gap density in lines per square inch. The final column of ampere 
turns is obtained by multiplying the ampere turns per inch by 
the length of each part. 

The total is 174.45 ampere turns for a pole flux of 45,000. 
The effective flux which cuts the armature inductors is some- 
what less than the air-gap flux and somewhat more than the 



48 



DESIGN OF SMALL MOTORS 



49 



armature-core flux. It will be about 39,250. This gives one 
point on the saturation curve of Fig. 17. The other points may 
be obtained in a similar manner by starting with other values 
of pole flux. In actual practice it is usual to check the curve 
experimentally by driving the motor as a separately excited 






AMPERE TURNS PER INCH SCALE B 
100 200 300 400 500 600 700 800 900 1000 1100 


1200 1300 


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/O 20 30 40 50 60 70 80 90 100 110 
AMPEI^E TURNS PER INCti SCALE A 

Fig. 22. Saturation Curves for Different Materials 



120 I5a 



generator at constant speed and reading the voltage for different 
field ampere turns. The voltage is then proportional to the flux. 
Returning to the design shown in Fig. 6, consider with the 
aid of the preceding tabulation why the proportion of the various 
parts is ideal. It is desirable to have the maximum flux through 



49 



50 DESIGN OF SMALL MOTORS 

the armature and the maximum slot section for the armature 
windings in order to secure the greatest power from this size of 
frame. If the diameter of the armature were increased, the slot 
section would be increased but there would be very little room 
for the field coils, causing a high current density in these coils 
and hence heavy losses, which would more than offset the gain 
due to increased slot area. The present diameter of 1| inches 
is best. We cannot use narrower teeth necks without increasing 
the teeth density so much that a considerable number of ampere 
turns will be required in the teeth. We might obtain more wind- 
ing space for the field coils and hence more ampere turns and 
flux by using a narrower pole section. The tabulation of ampere 
turns shows that the flux density in the pole core is needlessly 
low. But if the pole-core section were reduced, the field coils 
could not be form wound and slipped on, and hand winding is 
too expensive. We cannot reduce the yoke section without rais- 
ing the flux density to a point where the increased ampere turns 
required would exceed the additional ampere turns gained by the 
greater field-coil space. 

Air Gap. The air gap must be sufficient to give a comfort- 
able mechanical clearance between the armature and the field 
poles after the bearings have worn. Also the field ampere turns 
must be approximately 1.25 times the armature ampere turns in 
order to obtain good commutation. This i*equirement usually 
determines the air gap. 

Output Equation.* An equation which is useful in estimating 
the probable output from a proposed design of frame is 

j)2i ^ wattsX60.8Xl07 



r.p.m.X/g.XgX^ 

where Da is the armature diameter in inches; Lc is the frame 
length in inches; Mg is the mean pole-face density in lines per 
square inch (see Fig. 32) ; ^ is the ratio of pole arc to pole pitch; and 
q is the ampere conductors per inch of periphery (see Fig. 33). 
Heating. The heat-radiating capacity of the frame is also 
determined experimentally by winding the motor for any speed 
or voltage and measuring the electrical power input and the 

* See Gray's "Electrical Machine Design." 



50 



DESIGN OF SMALL MOTORS 51 

mechanical power output, the difference giving the losses. The 
value of watts loss which gives 50° C. rise may be taken as the 
maximum permissible watts loss, or the heat-radiating capacity, 
of the frame under test. The temperature rise is usually com- 
puted by measuring the change in field resistance according to 
the formula 

Rh = Ro (1 +0.004 T) 

where Ro is the coM resistance, Rh the hot resistance, and T the 
temperature rise in degrees centigrade. It is usually unnecessary 
to compute the local temperatures in the various parts of a 
standard frame, unless there are especially large losses in any 
one part, for example, on the commutator if the armature were 
wound for 4 volts and there were consequently heavy current 
and large brush contact and friction losses. In that case these 
losses would be computed and the commutator would have to 
be made longer , if the standard length gave more than 4 watts 
per square inch of commutator surface. 

Mechanical Design. The mechanical features of the design 
are similar to those shown in Fig. 2. The frame is made up of 
two castings. The main casing is a casting of iron turned 
inside to hold the field punchings. It is integral with the bearing 
at the commutator end. The end bracket at the pulley side is 
cast separately and fits into the turned rim of the main casing. 
The bearings have a bronze lining with wick oilers filled with 
vaseline. The brushes may be removed by loosening the side 
cap screws, which are threaded into a fiber bushing. 

In case the student should desire to construct a motor of 
this size and is unable to secure a second-hand frame for rewind- 
ing, the field structure may be cast of soft iron as a single unit 
similar in shape to the field punching given in Fig. 6, but with 
an outside diameter | inch larger to allow for the decreased per- 
meability of cast iron. The semicircular cuts at the top and the 
bottom may be omitted if desired. End brackets may be bolted 
to the frame, using any shape which will ensure accurate align- 
ment of the bearings. 

Another method of construction would be to cut a section of 
wrought-iron pipe about J inch thick and 2f inches more inside 
diameter. The poles may then be made from separate pieces of 



51 



52 DESIGN OF SMALL MOTORS 

cast iron machined to fit the inside of the pipe accurately. They 
may be held in place by machine screws threaded into them 
through the outside of the pipe. 

The armature should be made up of sheet-iron laminations. 
Circular discs may be cut out and circular slots employed which 
have been formed by drilling holes in one lamination at a time, 
the lamination being clamped between heavy templates which 
have been accurately laid out. The slot openings may be made 
by sawing into the holes after assembling the punchings. The 
diameter of the slot holes should be as large as possible without 
forcing the flux density in the necks of the teeth to a point where 
too many ampere turns are lost as computed by the method 
illustrated in the preceding tabulation. Li this case with a IJ- 
inch armature the 11 holes could be made J inch in diameter. 
Before assembly the laminations should be given a thin coating 
of shellac on one side. 

The commutator may be formed from a solid copper ring 
having clamping grooves cut in the sides. The ring should be 
accurately marked and cut into the required number of bars 
(22). The bars may be insulated with mica or micarta. 

i=HORSEPOWER MOTOR 

See Case I under Changing Characteristics of Motor by 
Rewinding. 

^"HORSEPOWER 1IO=VOLT COMPOUND MOTOR 

Requirements. The following motor might be used to drive 
a printing press : 



Requirements 



Output: ^ hp. at 1700 r.p.m. Must start under load when 

thrown on line without starting box. 
Input: 110 volts d.c. supply. Efficiency at full load not less 
[ than 70 per cent. 



A table of frame sizes of the Western Electric type SI) 
motors is given in Fig. 1. Previous experience with this line of 
frames indicates that the No. 264 frame will be best for this 
case. If the size smaller frame were tried, it would not be quite 
satisfactorv. 



52 



DESIGN OF SMALL MOTORS 



53 



Field Winding. From the saturation curve, Fig. 23, select 
an effective flux of 385,000 lines, which is about the right point 
near the knee of the curve and requires 900 ampere turns at no 
load. Allow L25X900, or 1125 ampere turns at full load. In 
order to enable the motor to start under full load, it will be 
necessary to employ a compound field winding. Supply 20 per 
cent of the full-load ampere turns with the series field, and 80 
per cent with the shunt field. This will give a strong field at 
starting without seriously affecting the speed variation, with 
variation in load under running conditions. 



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200 AOO too 800 1000 1200 1400 1600 1800 
M.M.F. IN AMPERE TURNS PER POLE 

Fig. 23. Saturation Curve for Type SD, Frame 264 Motor, at No Load 



Shunt Field. The shunt field ampere turns will thus be 
0.80X1125, or 900. From Fig. 5 it is seen that the section avail- 
able for the field winding without crowding will be about l|Xl, or 
1.125 square inch. Allow 80 per cent of this for the shunt winding, 
or 0.80X1.125, which equals 0.9 square inch. Assume that 
one-half of this section will be the total copper cross-section of 
the coil, or 0.50X0.9, which equals 0.45 square inch. The cur- 
rent density will be — --, or 2000 amperes per square inch. From 

Fig. 18 this corresponds to 3.05 watts loss per cubic inch. The 
length of mean turn of the field coil obtained from Fig. 5 will be 
twice the sum of the frame length, the pole-core width, and the 
corner allowance, which equals 2 (2.5+34-2.25), or 15.5 inches. 



53 



54 DESIGN OF SMALL MOTORS 

The volume of copper per coil will thus be length times 

section, or 15.5X0.45, which equals 6.98 cubic inches. The 

watts loss for two coils will be 2X6.98X3.05, or 41.85 watts. 

The shunt field current will be 41. 85 -^ 110, or 0.38 ampere. The 

'^8 
section of wire will be — — -, or 0.00019 square inch. No. 26 is 
^uuu 

0.0001996 square inch, which is near enough. Upon referring to 

Fig. 19 it is found that, by using enamel-covered wire, a space 

factor of about 0.55 can be obtained not allowing for taping. 

Since the section available for the shunt winding will be 0.9 square 

9x0 55 
inch, the number of turns will be ^' ^ • '^^ , or 2480. The resist- 

0.0001996 

ance of the two coils will be 2X2480X^X0.04563, or 292.5 

ohms. On 110 volts the current will be , or 0.376 ampere, 

and the w^atts loss will be 0.376X110, or 41.4 watts. The ampere 
turns will be 0.376X2480, or 933, which is 33 in excess of the 
requirements. The decision whether to connect the field long 
shunt or short shunt may be made later. So far it appears 
preferable to make it short shunt, thereby reducing the field cur- 
rent so as to get nearer the estimate of 900 ampere turns. 

Series- Field. The series field is to produce 0.20X1125, or 225 
ampere turns, with a coil section of 0.225 square inch. As the 
size of wire will be relatively large, assume a space factor of 0.6, 
or a copper section of 0.6X0.225, or 0.135 square inch. The cur- 
rent is not yet known, but an estimate may be made on the 
assumption of 75 per cent efficiency for the motor. Watts input 
will be Jx746-^0.75, or 498 watts, and since the voltage is to 
be 110, the current will be roughly 5 amperes. On this basis the 

225 
number of turns will be -— -, or 45 turns. The section of wire 

5 

will be ' , or 0.003 square inch. The wire nearest to this is 
45 

No. 14, which is 0.003228 square inch, giving _MM_^ or 42 

U.UUo^^O 

turns. The resistance of two coils will be 2 X 42 X—^X 0.002823, 



54 



DESIGN OF SMALL MOTORS 55 

or 0.306 ohm. The IR drop will be 5X0.306, or L53 volts, and 
the watts loss will be 5XL53, or 7.65 watts. This will be checked 
after the armature current has been determined more accurately. 
Armature Winding. If allowance is made for an armature 
IR drop ©f 7 volts, the back-e.m.f. will be 110 — 7 (armature IR) 
-1.5 (field IR)-2 (brush drop), or 99.5 volts. The number of 
inductors is found by means of the following equation: 

,, eXQOXcXW 99.5X60X2X10« nio • ^ + 
^^¥ ^r.p.m.X2P = 3850 00Xl700x2 = ^^' ^"^^^^^" 

912 
As there are 25 slots, there will be — — -, or 36.5 inductors per slot, 

zo 

which will be considered 36 inductors per slot. 

From Fig. 5 the gross slot area will be 0.1123 square inch. 

Allowing for a 0.02-inch slot fining and a 0.06-inch slot wedge at 

the top, the net area will be 0.0776 square inch. Assuming a 

space factor of 0.45, the copper section wiU be 0.45X0.0776, or 

0.035 square inch per slot. The section of each inductor will be 

5^, or 0.000972 square inch. No. 19 is 0.001012 square inch, 

which is sufficiently close, as the space factor for this size of wire 
is above 0.45. Use a simplex singly re-entrant winding corded 
2J slots, that is, a pitch spanning 10/25 of the entire number of 
slots. The length of mean inductor will be the sum of the length 
of the armature and the length of one end connection, which 
equals 2.5+5.5, or 8 inches. The total length of armature wire 

will be :^X36X25, or 600 feet. The resistance per foot of No. 19 
iz 

wire is 0.009 ohm (hot), and the resistance of the armature with 
two paths in parafiel of 300 feet each wiU be J X 300X0.009, or 
1.35 ohms. 

Losses. First determine the ^various losses and then the arma- 
ture current. The volume of iron in the teeth necks will be 
25X0.19X0.45X2.5, or 5.34 cubic inches, and the weight wifi be 
5.34X0.278, or 1.473 pounds. There wiU be eight teeth under one 

pole at a time, and thus the tooth densitv will be , 

^ ^ 8X0.19X2.5' 

or 101500 lines per square inch. Since the revolutions per minute 



55 



56 DESIGN OF SMALL MOTORS 

1700 
are to be 1700, the frequency in the armature will be -^tt-, 

or 28.33 cycles. Use annealed sheet iron 0.014 inch thick for the 
punchings. Fig. 21 gives 6.5 watts per pound at 101,500 lines per 
square inch at 28.83 cycles. The iron losses in the teeth will 
therefore be 6.5X1.473, or 9.57 watts. In the armature core the 
volume of active iron will be 2.5X7r (1.65^-0.5^), or 2.5X7rX2.47, 
which equals 19.41 cubic inches, and the weight will be 19.41 X 

0.278, or 5.39 pounds. The flux density will be — --, 

JXl. 1X^.5 

or 70000 lines. From Fig. 21 the loss per pound will be 3 watts. 
The iron losses in the core will thus be 3X5.39, or 16.16 watts. 
The total iron losses will be 9.57+16.16, or 25.73 watts. The 
losses per pound given in Fig. 21 are large enough to cover any 
stray power losses in a motor of this size. 

The commutator friction loss will depend upon the brush 
area, and this, in turn, will depend upon the armature current, 
which is not yet determined. It may, however, be assumed for 
this purpose that it will not differ materially from the estimate of 
5 amperes. Use grade E-1 ^'Le Carbone" carbon brushes, which 
are good for 40 amperes per square inch, giving a minimum area 
of -io, or I square inch per brush, and choose a brush f X|, or 
0.141 square inch in area, and having a spring pressure of 2.5 
pounds per square inch. The coefficient of friction will be about 
0.3. As the commutator is to be 3.5 inches in diameter, the 
peripheral speed will be 3.5X7rXl700^12, or 1560 feet per min- 
ute. The friction loss for two brushes will be 2X2.5X0.141 X0.3X 

1560, or 330 foot pounds per minute, which equals , 

or 7.46 watts. 

The bearing loss will be 0.8 dl ( — — j watts,* where d is bear- 
ing diameter in inches, I the length in inches, and v the rub- 
bing velocity in feet per minute. By substituting in the formula, 
the loss for both bearings is found to be 2X0.8X0.75X3.5X 
0.75X7rXl700\^ 



12X100 J 



or 25.6 watts. 



See "Standard Handbook for Electrical Engineers," 4th ed., Sec. 8, H 140. 

56 



DESIGN OF SMALL MOTORS 57 

The windage loss may be neglected as it will be very small 

and is difficult to compute. 

The armature input (less the PR loss) will be as follows: 

Output in mechanical power 746 -^ 2 373.00 watts 

Iron and stray power losses 25.73 watts 

Brush friction 7.46 watts 

Bearing loss 25.60 watts 

Total 431.79 watts 

Using the original estimate of 99.5 volts back-e.m.f., the current 

will be , or 4.34 amperes. Since the resistance of the 

99.5 

armature was found to be L35 ohms, the armature IR drop will 

be 4.34X1.35, or 5.86 volts. Using a short-shunt connection, the 

series field current will be 4.34+0.371, or 4.711 amperes, and the 

series field IR drop will be 4.711X0.306, or 1.442 volts. Thus 

the corrected value of back-e.m.f. will be 110 — 1.442 — 5.86 — 2, or 

100.7 volts. The corrected value of armature current will be 

431.79 , ^^ 

, or 4.29 amperes. 

The corrected value of ampere turns will be: shunt field, 

— — ^ — X2480, or 921 ampere turns; and series field, 4.711X42, 

292.5 ohms 

or 198 ampere turns, the total field ampere turns per pole being 

1119 
1119. The corresponding effective ampere turns will be , 

i.zo 

or 895. From Fig. 23 the effective flux will be 383,000 lines. It 

has been decided to use 36 inductors per slot, and thus every 

factor in the voltage equation has been determined accurately 

except the speed. This will be: 

__ eXmXcXW ^ 100.7 X60X2XW ^ - 
^'^'^' NX^X2P 36X25X383000X2 
which is sufficiently close to 1700. Since there are to be 36 
inductors per slot, 25 slots, 2 inductors per turn, and 2 poles, 
and the armature current is to be 4.29 amperes, the armature 

ampere turns per pole will be ' — - — - — ^ — , or 483. The ratio of 

z x^xz 

field to armature ampere turns per pole will be ^^r-, or 2.315, 
which is over 1.25 and hence satisfactory. 



57 



58 DESIGN OF SMALL MOTORS 

In the case of i-hp. motors of 110 volts or higher and all 
larger sizes it is necessary to check up another feature of the 
design, which affects commutation, and is called the reactance 
voltage. This voltage is 

er = KS (r.p.m.) LLcT^f ^^\o-' volts 

where S is commutator bars, 7c is armature inductor current, Lc 

is frame length in inches, T is turns per coil between bars, and K 

is a constant between 0.93 and 1.6. Substituting the values found 

for this de'sign gives 

e, = 1.6x50Xl700x2.145X2.5X8lXlXl0-8 = 0.59 volt 

er should be less than to of the voltage drop per pair of brushes, 

which is true in this case. 

The various losses will be as follows: 

Shunt field copper 0.371 X 108.6 40.30 watts 

Series field copper 4.712X0.306 - 6.77 watts 

Armature copper 4.292 X 1.35 24.80 watts 

Brush contact 2X4.29 8.58 watts 

Iron and stray power loss 25.73 watts 

Brush friction 7.46 watts 

Bearings and windage 25.60 watts 

Total 139.24 watts 

lN/7. 

Summary. The efficiency will thus be 



iX (746) + 139.24' 

which equals 0.728, or 72.8 per cent, which meets the original | 

requirements. The heating will be well within permissible limits i 

as the losses will be well distributed and will total only 139.24 ! 

watts, while the heat-radiating capacity for 50° C. rise for this i 
frame. No. 264, is 175 watts. 

The shop information for this 110- volt, d.c, J-hp. motor will | 
be as follows: 

Mechanical Data Frame No. 264, type SD. Brushes grade E-1 "Le | 

Carbone" fXf section with 5^ ounces spring pressure, j 

All other parts standard with No. 264 frame. | 

Electrical Data \ 

Shunt Field Winding: 2480 turns per coil of No. 26 single cotton-covered wire, j 

Coils connected in series, short shunt. | 

Series Field Winding: 42 turns per coil of No. 14 double cotton-covered wire, i 

Wrapped with 0.03-inch tape. i 

Armature Winding: 36 inductors per slot of No. 19 double cotton-covered j 
wire. Simplex, singly re-entrant, 10/25 pitch. Standard 

slot fining. ' 



58 



i 



DESIGN OF SMALL MOTORS 59 

INDUCTION MOTOR DESIGN 

Characteristics of Induction Motor. The complete treatment 
of induction motor design is beyond the scope of this article, but 
the following abridged discussion will give practical results in the 
case of small motors which will not differ materially from the 
results obtained by the longer and more complete methods 
which are generally employed in the case of large motors. The 
induction motor consists of two main members; a primary mem- 
ber, which may be the stationary or the revolving member; and 
a secondary member, which in small motors is usually furnished 
with a squirrel-cage winding consisting of copper bars and end 
rings. When the motor is running at normal speed the resultant 
effect of the primary and secondary currents is to produce a 
practically steady field of flux* which rotates with respect to the 
primary member, just as the flux is steady and may be consid- 
ered as rotating relative to the armature in a d.c. motor. In a 
d.c. motor the field poles are salient physical poles, while in an 
induction motor there are no physical poles but the flux is dis- 
tributed as though there were poles, except that this configuration 
of flux does not maintain a fixed position with respect to either 
member but rotates slowly with respect to the secondary, or 
squirrel-cage member, and rapidly with respect to the primary 
member. 

In a d.c. motor the relative rotation between the field flux and 
the armature generates an a.c. back-e.m.f. in the armature coils, 
which is changed to d.c. back-e.m.f. by the commutator. Sim- 
ilarly in the induction motor the relative rotation of the flux 
generates an a.c. back-e.m.f. in the primary member. But as 
there is also a slow rotation between the flux and the secondary 
member, another a.c. e.m.f. is generated in it, the frequency of 
which is correspondingly slow and is called the slip frequency. 
Upon these premises, which are approximately true, the following 
simplified treatment is based. 

|=HORSEPOWER MOTOR 

Requirements. The first induction motor considered will be 
one giving an output of J hp. 

* Note the discussion given later under Magnetizing Current. 

59 



60 



DESIGN OF SMALL MOTORS 



I 



rOutput: I hp. at 1800 r.p.m. synchronous speed. Efficiency not 
Requirements ] less than 60 per cent. Power factor not less than 0.55. 

llnput: 110 volts, single phase, 60 cycles. 

The Western Electric type SA, form SSI, frame No. 137, 
Fig. 24, will be used in this design. The design of this frame 
will be treated later as a separate problem, it being assumed that 
the line of standard frames given in Fig. 25 is available. These 
frames are all intended to carry a main winding and a starting 
winding which is cut out by a centrifugal device. 

Sinusoidal Distribution of Flux. A disassembled view of the! I 
motor is shown in Fig. 8. As the synchronous speed is to be 

TYPE' 3 A. FRAME'N9 IZt 
LEN6TH-L87'f 

AIR 6AP =.012'* 




PRIMARY- 
36 SLOTS 
SLOT DEPTH ^.4-Sn 
AVE WIDTH ^J97f 
5L0T OPENING =. 10" 
TOOTH NECn ^ JO" 

SECONDARY' 
59 5L0TS 
5L0T DIA. - .I95'f 
SLOT OPENING^ .05f^ 



SCALE: HALF SIZE 

Fig. 24. Details of Western Electric Type SA, Frame 137 Motor | 

1800 r.p.m., or 30 revolutions per second, and the frequency 60 
cycles, there will be two cycles per revolution and hence two pairs 
of poles, or four poles. At any instant of time the flux will be 
distributed approximately sinusoidally. There will thus be four 
regions in the air gap where the flux density is zero and four 
where it is a maximum, and in the intermediate regions the 
flux density will be intermediate in value in the same propor- 
tion as the height of a sine wave. Referring to Fig. 26, con- 
sider the instant when the four regions of maximum flux density 
come under the four groups of teeth 5 and 6, 11^ and 15, 28 and 



60 



DESIGN OF SMALL MOTORS 



61 





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61 



62 



DESIGN OF SMALL MOTORS 



2J+, and S2 and SS. These two middle teeth in each quarter, 
such as 5 and 6 in the first quarter, . will have maximum flux 
density; teeth 4 and 7 will have somewhat lower density; teeth S 
and 8 still lower; teeth 2 and 9 still lower; and teeth 1 and 10 
zero density at the particular instant of time under consider- 
ation. The same will be true for each of the other three quar- 
ters. The main winding will be distributed as shown, except 
that the number of conductors in each slot will be multiplied by 
a factor Xi, which will be determined later. The starting wind- 
ing is shown in Fig. 27 and the number of conductors shown in 

JYPt SA. FQAMt N9I37 MAIN WINDIN6 




MULTIPLY CONDUCTORS IN EACH 5L01 
T0RIN6 BY Ki -21 TO PIN 6 

Fig. 26. Main Winding for Western Electric Type SA, Frame 137 Motor 



each slot will be multiplied by a factor K2. It will be noted 
that the coil centers of the starting winding will lie midway 
between the coil centers of the main winding, so that the motor 
will act as a two-phase machine when starting. 

Back=E.M.F. How to compute the value of the back-e.m.f. 
in the main winding should next be considered. The distribution 
of the windings will be arranged so as to secure a back-e.m.f. of 
nearly sine shape under the action of the distributed rotating 
flux previously described. It is seen that the turns in any one 
quarter of the winding will be concentric, and hence the electro- 
motive forces induced will be in phase and may be added directly, 



62 



DESIGN OF SMALL MOTORS 



63 



The flux in any one tooth will vary from zero to maximum value 
according to a sine function of time since the rotating field will 
be of sine distribution. But since the sum of any number of sine 
waves of any phase relation is also a sine wave, the total flux 
enclosed in any turn will be a sine function of time. Therefore, if 
the maximum value of flux enclosed in each turn (which will be 
the value when the wave crests coincide with the coil centers) 
is computed, the induced e.m.f. in the turn may be computed by 
the ordinary formula used in transformer design 



e = 



V^Tff^N 



W 



TYPE 5 A. FPAM£N9/57 5TAPT/N6 WINDING 




MULTIPLY CONDUCTORS IN EACH SLOT 
T0RIN6 BY K2 =/5 TO RIN6 

Fig. 27. Starting Winding for Western Electric Type SA, Frame 137 Motor 



where <^ is the maximum instantaneous value of flux, A" the num- 
ber of turns, and / the frequency. 

Assume the maximum mean flux density in the air gap to be 
25,000 Hues per square inch.* The tooth pitch will be 0.341 inch 
and the frame length L81 inches; therefore the number of lines 
through a tooth in the region of maximum density will be 25000 X 
0.341 XL81, or 15400. kt the instant when this maximum value 
of flux occurs in teeth 5 and 6, the flux in all the teeth will be as 
follows: 



• In general, the gap density is somewhat lower than in d.c. motors and r%nge8 from 20 OCKJ 
to 35,000 for motors up to 1 hp having partly closed slots. 



et 



64 DESIGN OF SMALL MOTORS 



Teeth Numbers 


Per Cent of Maximum 


Flux in Teeth 


5 and 6 


100 


15^.00 


4 and 7 


85 


13080 


3 and 8 


62 


9550 


2 and 9 


35 


5390 


1 and 10 








Total. . 


43420X2 



The voltages induced in one quarter of the main winding are 
given in the following tabulation: 







Flux Included 

(From previous 

tabulation) 


Flux XTurns 


Induced Voltage 


Turns 


Teeth Included 


^XN 


V27r/$iV10-8 


Two outer 


2,3,4,5,6,7,8,9 


86840 


173680 


0.4640 volt 


Next two 


3,4,5,6,7,8 


76060 


152120 


0.4060 volt 


Next one 


4,5,6,7 


57960 


57960 


0.1548 volt 


Inner turn 


5;6 

Total 


30800 


30800 


0.0822 volt 




...414560 


1.1070 volta 



4 



The total voltage is to be multiplied by four for the four quarters 
and by a factor Ki, by which the number of conductors per slot 
shown in the diagram is to be multiplied. If the back-e.m.f. is 
estimated as 93 volts, leaving the vector difference between 110 
volts and 93 volts as the amount of impedance drop in the wind- 
ing, the following relation is true: 

1.107x4Xi^i = 93 volts 
and therefore 

Xi=_-^^ = 21.05 
1.107X4 

If Ki is made to equal 21, there will be a back-e.m.f. of 1.107 X 
4X21, or 92 volts. 

Main Winding. There will thus be 2Ki, or 42 conductors 
per slot, in the main winding in those slots in which 2 conductors 
are shown on the diagram. The total number of conductors in 
the main winding will be 48X21, or 1008. In order to obtain the 
best efficiency, as large a size of wire as can conveniently be 
wound in the slot will be used. The slot area. Fig. 24, will be 
approximately 0.45X0.197, or 0.0886 square inch. Allowing for 
a 0.01-inch slot lining and a slot wedge at the top 0.03 inch thick, 
0.07 square inch will be left for winding space. Assume that 0.5 
of this will be copper section, or 0.5X0.07, which equals 0.035 

square inch. The size of conductor will thus be — , or 0.000833 

42 



64 



DESIGN OF SMALL MOTORS 65 

square inch. No. 20 is 0.0008023 square inch, which is sufficiently 
close; and No. 20 will therefore be used for the main winding. The 
length of mean conductor will be approximately the length of a 
conductor in the slot between teeth 2 and S. The end connection 
will extend over six teeth, which distance, when laid off on Fig. 24 
at the center of the slots, is L7o inches. The active length of 
the conductor will be L81 inches, but | inch at each end must 
be allowed, making the total length of mean conductor L754- 
0.5+L81+0.5, or 4.56 inches, or 0.38 foot. The total length of 
wire will be 0.38X1008, or 383.5 feet. The resistance per foot 
will be 0.01135 ohm, and the total resistance will be 383.5 X 
0.01135, or 4.35 ohms. Before proceeding further it will be well 
to estimate the magnetizing current and the power factor. 

Magnetizing Current. The total ampere turns per pole are 
computed in the same way as for a d.c. machine, and the neces- 
sary calculations are given in the following tabulation. Fig. 22 
is used to obtain the ampere turns per inch. 

Ampere Length Ampere 

Flux Turns of Part Turns 

Part Flux Section Density per Inch (in.) per Pole 

Primary core 86840-^2 0.5X1.81 47900 5.0 1.25 6.25 

One tooth 15400 O.lXl.81 85100 24.5 0.45 11.03 

Air gap 15400* 0.2435X1.81 35400 11080.0 0.012 133.00 

Secondary core 80000^2 0.5X1.81 44250 4.5 1.5 6.75 

Total 157.03 

The value of the flux density in the air gap which is used to 

compute the ampere turns is greater than the crest value of the 

mean air-gap density owing to the slot openings. In machines 

with open slots it is necessary to employ a semi-empirical formula 

in order to obtain sufficiently accurate results. In machines of 

1 hp. or less it is the practice to employ nearly closed slots. In 

such machines the slot contraction factor may be taken as 

TF<+2A 

, where Wt is the width of the tooth; A, the air-gap length; 

and p, the slot pitch. This contraction factor must be found for 
both primary and secondary members, and the combined contrac- 
tion factor will be the product of the two. The combined factor 
multiplied by the slot pitch gives the part of the tooth pitch which 

* Under one tooth at maximum density. 



65 



66 DESIGN OF SMALL MOTORS 

is effective as a flux path. In this case the primary contraction 

.„ , 0.241+0.024 ^^_ , ^, , 

factor will be r— — j , or 0./7/, and the secondary contrac- 

tion factor ^'^^^^"t^'^^^ ov 0.92. The slot pitch is to be 0.341 
0.3166 

inch. The effective part of the primary tooth pitch will thus be 

0.777X0.92X0.341, or 0.2435 inch. 

Returning to the value of ampere turns, 157.03, it is neces- 
sary to multiply this by a factor w^hich accounts for the part of 
the excitation supplied by the secondary winding during the time 
that the current in the primary is below peak value. If the 
rotating field were of uniform value this factor would be 2, and 
in large single-phase motors it is almost 2. In small motors it 
varies from 1.8 for a |-hp. motor to 1.9 for a 1-hp. motor. The 
ampere turns per pole will be 1.8X157.03, or 282.5 at the instant 
of maximum flux and hence maximum magnetizing current. The 
effective value of the magnetizing ampere turns per pole will thus 
be 282.5 -^'^/2, or 200. The number of turns per pole on the 
main winding will be 126, and hence the magnetizing current will 
be 2004-126, or 1.587 amperes. 

In order to judge whether 1.587 amperes is a reasonable 
magnetizing current, assume that the design is to have an effi- 
ciency of 60 per cent* and a power factor of 0.55. The total 

746 X- 
current will then be — — — ^ , or 2.57 amperes. 

The power component of the total current will be 0.55X2.57, or 

1.41 amperes, and the reactive component will be V2.d7^ — 1AV, 

which equals v4.61, or 2.15 amperes. A magnetizing current of 

1 587 
1.587 amperes will be '^, , 0.74 of the total reactive current. 
2.15 

This ratio should vary from 0.85 in the very smallest motors to 

0.6 in large motors, depending upon the air gap and the flux 

density used, etc. Another useful check is the number of in- 

phase component ampere conductors per inch of periphery, which 



* The eflBciency of induction motors will vary in the following manner: around 55 per cent 
for a i^i-hp. motor, aroimd 70 per cent for a i-hp. motor, and around 80 per cent for a 1-hp. motor. 
The corresponding range for power factor is 0.50 to 0.85. In general, the eflBciency will be higher 
as tho frequency increases and the number of poles decreases. 



66 



(DESIGN OF SMALL MOTORS 67 
hould be less than 200 for motors under J hp. In this case it is 
, or 116. The design may now be proceeded with. 
0.9 Xtt 
Starting Winding. In the starting winding the current should 
be as nearly in phase with the voltage as possible and at the same 
time develop the maximum number of ampere turns. A good 
compromise will be obtained if the number of conductors is made 
one-half the number used in the main winding, or 504. The 
winding is conveniently distributed as in Fig. 27, in which 40, 
the number of conductors shown, should be multiplied by a 

504 
constant K2 to give the total number. ^2 will thus be — — , or 

12.6. Make K2 equal 13 and the total number of conductors 520. 
The resistance of the starting winding must be made high enough 
to limit the current drawn on starting to a satisfactory value. 
In such small motors a reasonable value for the current at start- 
ing, through the starting winding only, is four times the normal 
full-load current, which in this case will be 4x2.57, or 10.28 

amperes. The impedance of this winding will thus be -— — , or 

10.7 ohms. Since the larger part of this impedance will be resist- 
ance, the resistance of the starting winding may be taken as 
10 ohms. The total length of mean conductor, which is found in 
the same way as that for the main winding, will be about 5.25 
inches, or 0.4375 foot; it will be somewhat greater than that for 
the main winding, as the starting winding will be wound on over 
the main winding. The length of wire will thus be 0.4375X520, 

or 227.5 feet. The resistance per foot will be , or 0.044 ohm. 

227.5 

The resistance of No. 26 is 0.04081 ohm per foot (cold), and No. 26 will 
therefore be used for the starting winding. The total copper section 
for the 26 conductors per slot will be 26X0.0001996, or 0.0052 square 
inch. This is considerably less than one-half the copper section of 
the main winding, so that there will be ample room in the slots 
which contain conductors partly of the main and partly of the start- 
ing winding, for example, the slot between teeth 3 and 4- 

Secondary Winding. The secondary member will be wound 
with a squirrel-cage winding completely filling the slots, which 



87 



68 DESIGN OF SMALL MOTORS 

will be circular and 0.195 inch in diameter. The end ring should 
have sufficient section so that the current density in it will be 
about the same as in the bars, which requires the end-ring section 
to be about one-third of the total section of the bars per pole, or 

_Lx— xO.lOo^X— , which equals 0.0969 square inch. In the 
3 4 4 

No. 137 frame the end ring is of rectangular section and is used 

to support the end shields, which are turned to fit over the outer 

edge, Fig. 9. The width is therefore made 0.56 inch, which gives 

0.0969 
the proper seating. The thickness will be ' , , or 0.173 inch. 

0.50 

The volume of copper in the secondary winding will be: 39 bars, 

39X— xO.i952xl.8l, or 2.1 cubic inches; and 2 end rings, 2X 

0.0969 X4.6X7r, or 2.8 cubic inches. 

Losses. The iron losses will be divided into hysteresis loss 
and eddy-current loss. The volume of iron in the tooth necks will 
be 36X0.1X0.45X1.81, or 2.93 cubic inches. The maximum flux 
density /8„, in the teeth was found to be 85,100. As there are 
losses in the tooth tips due to pulsation of flux and losses due to 
short-circuited laminations, the following empirical formula is 
given by Hobart for 0.014-inch laminations at 60 cycles: 

Watts per cu. in. =4.1 f^X 

Therefore in this case the teeth losses will equal 4.1X0.85PX2.93, 
or 8.88 watts. The volume of active iron in the primary core will 
be 0.5X1.81 X2.4X7r, or 6.82 cubic inches. The flux density was 
found to be 47,900 lines per square inch. The iron loss will 
therefore equal 4.1X0.4792X6.82, or 6.4 watts. The iron losses 
in the secondary member will be negligible on account of the low- 
frequency. The sum of the iron losses will therefore be 8.88+6.4, 
or 15.28 watts. 

The bearings are to be ^ inch in diameter and IJ inches long. 

The bearing losses for two bearings will be 2X0.8 dlf—j\ 
which equals 2X0.8x0.5xl.5x(i^^5^^^^^ or 4 watts, 



68 



DESIGN OF SMALL MOTORS 69 

where d and I are the bearing diameter and length in inches, 

respectively, v is the rubbing velocity in feet per minute, and 

1700 r.p.m. is assumed as a tentative speed. 

The brushes which bear on two slip rings attached to the 

primary member as shown in Fig. 9 will have a section of iXj, 

or 1^ square inch, and a pressure of 2.5 pounds per square inch, or 

2 5 

--^, which equals 0.1562 pound. The diameter of the rings will 

be 2.75 inches, so that the power loss due to brush friction, 

2 7'i 
assuming a coefficient of 0.3, will be 2X0.3 X 0.1562 X-f^r-XTrX 1700, 

115 
or 115 foot pounds per minute, which equals X746, or 2.6 

ooUUU 

watts. Allowing a brush drop of 2 volts, the brush contact loss 

will be 2X2.6, or 5.2 watts. The windage loss will be very small, 

but 1 watt may be allowed for it. 

The PR loss in the secondary may be estimated by finding 
the component of current in the primary which corresponds to 
the total mechanical power output, adding to this the part of the 
magnetizing current transformed to the secondary, and assuming 
that there is an equal number of secondary ampere turns. 

The total mechanical output of the primary will be as 
follows: 

Power delivered by motor i X746 93.3 watts 

Bearing loss 4.0 watts 

Brush friction loss 2.6 watts 

Windage loss 1.0 watt 

Total 100.9 watts 

The back-e.m.f. of the primary is to be 92 volts, so that the 

desired component of current in the primary will be , or 1.1 

amperes. The total value of the magnetizing current was found 

to be 1.625, of which ^X 1.625, or 0.723 ampere will be deliv- 
1.8 

er ed to the s econdary, so that the total transformed current is 
Vl.P-j-0.723^ or 1.317 amperes. These components are nearly, 
though not quite, at right angles. Since the number of con- 
ductors per pole is to be 252, there will be 1.317X252, or 332 



69 



70 DESIGN OF SMALL MOTORS 

ampere conductors per pole. There will be a like number of 
ampere conductors per pole in the secondary. The current density 
will vary from bar to bar in approximately a sine wave propor- 
tion, so that the maximum ampere conductors per bar will be 

TT 1 

found by plotting to be — X^^X332, or 53.4 amperes per bar, 

53 4 
which will give a current density of , or 1795 amperes 

-^XO.1952 
4 

per square inch. From Fig. 18 this corresponds to 2.47 watts per 

cubic inch. The volume of copper in the bars was found to be 

2.1 cubic inches, making the secondary watts 2.47x2.1, or 5.19 

watts. The current in the end rings will be 332 -^ 2, or 166 

amperes, which will give , or 1713 amperes per square inch, 

u.uyuy 

or, from Fig. 18, 2.23 watts per cubic inch. The volume of 
copper in the bars was found to be 2.8 cubic inches, giving 
2.23X2.8, or 6.24 watts. The total secondary copper loss will 
therefore be 5,19+6.24, or 11.43 watts. The primary copper loss 
can only be approximated until the total primary current has been 
determined. Using the previously estimated value of 2.57, the 
primary PR loss will be 2.57^X4.35, or 29.34 watts. 

Slip. The secondary copper loss may be used as a means of 
determining the slip. In a polyphase motor the slip is directly 
proportionate to the total secondary copper loss. In a single- 
phase induction motor, however, the secondary current is not in 
phase with the flux and hence, in determining the slip, it is 
necessary to consider only that part of the secondary loss which 
is due to the in-phase component of the secondary current. The 
ratio of this current to the total secondary current was found to be 

' , and the loss due to the in-phase current in the secondary 

will be ("-y-Yx 11.43, or 7.99 watts. The slip will thus be 

7.99 

^ X1800 r.p.m., or 141.8 r.p.m., and the speed of the motor 

at full load will be 1658 r.p.m., which is satisfactory for such a 
small motor. It may be said that a higher secondary loss will 



70 



DESIGN OF SMALL MOTORS 71 

give better starting torque, and hence it is frequently advisable to 
design the motor with higher loss and thus greater slip. 
The total losses will be as follows: 

Primary PR loss 29.30 watts 

Secondary PR loss 11.43 watts 

Iron losses 15.28 watts 

Bearings and windage 5.00 watts 

Brush contact loss 5.20 watts 

Brush friction 2.60 watts 

Total 68.81 watts 

Summary. The output of the motor is to be 1x746, or 93.3 

93 3 93 3 

watts, so that the efficiency will be , ^o qi > ^r ' , which 

yo.o-rbo.ol Ib^.ll 

equals 0.575, or 57.5 per cent, which is somewhat lower than the 

original estimate of 60 per cent. The corrected value of the primary 

power component of the current will be 162, or 1.471 amperes. 

It is not possible to compute the power factor without finding the 

impedance of the primary and the secondary windings. It may be 

assumed, however, as explained in the previous calculation of the 

magnetizing current, that the magnetizing current will be about 

80 per cent of the total reactive component of current and will, 

therefore, be ' , or 2.03 amperes. The total primary current 
O.oO 

will then be V2. 03^-1-1. 47 P, or 2.51 amperes, which is close to 

the first estimate, so that the efiiciency need not be corrected. 

1.471 
The power factor will thus be ' , or 0.587, which is satisfac- 

^.ol 

tory for such a small single-phase motor. 

It is not possible to check up the assumption that the back- 

e.m.f. of the primary will be 92 volts without computing the 

impedance of the primary winding. This is complicated to compute 

and would not affect the determination of the windings, which is the 

main purpose of this treatment. It will be sufficient therefore to 

note that the primary IR drop is 2.51X4.35, or 10.92 .volts, 

which is less than the difference between 110 and 92, and as the 

impedance drop is usually within one and one-half to three times 

the IR drop and is added at a vector angle to the back-e.m.f., it 

may be assumed that the design will give satisfactory operation. 

The practical effect of any discrepancy will merely be to vary the 



71 



72 



DESIGN OF SMALL MOTORS 



flux density a small amount one way or another. In general, in 
working out a design with the above approximate method it will 
be satisfactory if the primary IR drop is about one-half the direct 
difference between the back-e.m.f. and the impressed e.m.f., pro- 
vided the recommendations as to ampere conductors per inch of 
periphery have been followed. 

The computation of the starting torque is beyond the scope 
of this treatment, but if the rules given are followed it may be 
assumed that the starting torque will be what is ordinarily con- 
sidered satisfactory. 

The approximate way to estimate the heating of a motor on 
which no test data is available and in which the losses are rea- 



to 0.014 

%Uaoi2 

^^0.010 
^"^0.008 
^^0.006 
S^ 0.004 
fig 0.002 

i 



TEMPERATURE RISE FOR ENCLOSED MOTORS 




RADIATING SURFACE 



:^ 



12 5 4. X/0^ 

ARMATURE PERIPHERAL VELOCITY IN FT PER MIN. 

Fig. 28. Temperature Rise for Enclosed Motors 
Courtesy of " Standard Handbook for Electrical Engineers," ^th ed., page i 



a 



sonably well distributed is to employ the formula given in Fig. 28 
and compute the temperature rise for a completely enclosed motor. 
It may then be assumed that the actual temperature rise will be 
from 40 to 80 per cent of this, depending upon the degree of 
ventilation secured. In this case the peripheral velocity will be 

— — — — — , or 1630 feet per minute, which will give 0.0078 watt 

per square inch for 1°C. rise. The radiating surface will be 

or 7r5.35( -^- — 1-4 ), which equals 112.3 square inches. 



7r2)(|+I 



72 



DESIGN OF SMALL MOTORS 73 

With 68.81 watts loss, the rise, if the motor were enclosed, would 

68 81 
be , or 78.4° C. As this design is to be fairly well 

ventilated and is to have a small punched fan, it may be assumed 
that the rise will not exceed 60 per cent of this, which equals 
0.6X78.4, or 47° C, which may be considered satisfactory. 

The shop information on the f-hp., 110- volt, single-phase 
induction motor will be as follows: 

Mechanical Data Frame No. 137, type SA. 

Electrical Data 
Main Winding: As per Fig. 26. 1008 conductors of No. 20 double 

cotton-covered wire. 
Starting Winding: As per Fig. 27. 520 conductors of No. 26 double cotton- 
covered wire. Wind over main winding and connect in 
parallel with same through centrifugal cutout to slip 
rings. 
Secondary Winding; Standard end rings, 0.56X0.173 section. All other 
details standard with No. 137 frame. 

i=HORSEPOWER MOTOR 

See Case VI in Changing Characteristics of Motor by Rewind- 
ing, under the head Induction Motors. 



Requirements. The student is referred to the previous design 
of the 1-hp. induction motor for a fuller explanation of the various 
steps than is given in the treatment of this |-hp. motor. 

(Output: I hp. at 1800 r.p.m. synchronous speed. Efficiency not 
less than 65 per cent. Power factor not less than 0.70. 
Input: 110 volts, single phase, 60 cycles. 

The Western Electric type SA, form SI, frame No. 165, Fig. 29, 
will be employed. 

Distribution of Flux. The synchronous speed is to be 1800 
r.p.m., or 30 revolutions per second. This is two cycles per revo- 
lution and requires two pairs of poles, or a four-pole winding. 
In order to obtain a good distribution of the main and the start- 
ing windings, the main winding will be as shown in Fig. 30. 
Assume a crest mean flux density in the air gap of /3 equaling 

25,000 Hues per square inch. The tooth pitch will be -^— - — , or 



73 



74 



DESIGN OF SMALL MOTORS 



0.413 inch, and the length of the armature will be 2.5 inches, 
so that the flux through a tooth in the region of crest density 
vnW be 0.413X2.5X25000, or 25800 lines. 

TYPE-SA. FORM- 51. 

FRmt-m 

FRAME LENGTH = 2,5" 
AIR 6AP =.012" 

PR/MARY- 
48 5L0T5 
5L0J DlPJtl = 75" 
AVE WJDW - .203" 
SLOT OPENING = .125 '< 
TOOTH NEC/\ - .150" 

SECONDARY- 
37 5L0T5 
5L0T DJA -30" 
5L0T OPENING = .06" 

SCALE QUARTER 5/ZE 

Fig. 29. Details of Western Electric Type SA, Form SI: Frame 165, Single-Phase 
Induction Motor 



li 




TYPE 5 A. FORM SI. FRAME 165. 
MAIN WINDING 




Fig 30. Main Winding for Western Electric Type SA, Form SI, Frame 165, Single-Phase 

Induction Motor 

Consider the instant when the center of one of the four 
regions of maximum flux density comes under tooth 7, which is 
the center of one of the four groups of turns. The flux through 
the teeth in the first quarter of the winding will be as follows: 



74 



I 



DESIGN OF SMALL MOTORS 



75 



Teeth Numbers 


Per Cent of Maximum 


Flux in Teeth 


7 


1.0000 


25800 


6 and 8 


0.9659 


24950 


Sand 9 


0.8660 


22350 


4 and 10 


0.7071 


18230 


3 and 11 


0.5000 


12900 


2 and 12 


0.2588 


6680 


1 and 13 


0.0 





Total.. 


110910X1 



Back=E.M.F. The following tabulation shows the voltages 
induced in one quarter of the main winding: 



Turns 

Two outer 
Next two 
Next two 
Next two 
Next one 
Inner turn 


Teeth Included 

2,3,4,5,6,7,8,9,10,11,12 

3,4,5,6,7,8,9,10,11 

4,5,6,7,8,9,10 

5,6,7,8,9 

6,7,8 

Total 


Flux Included 

(From previous 

tabulation) 

196220 
182860 
•157060 
120600 
75900 
25800 


Flux XTurns 

392440 
365920 
314120 
241200 
75900 
25800 


Induced Voltage 
V27r/$iVlO-8 
1.0470 volts 
0.9760 volt 
0.8390 volt 
0.6430 volt 
0.2025 volt 
0.0689 volt 




. . 1415380 


3.7764 volts 



The total back-e.m.f. for the four quarters will thus be 3.7764 X 
4:XKi, where Ki is a factor to be determined by which the number 
of conductors in the wiring diagram is to be multiplied. If a 
back-e.m.f. of 100 volts out of the 110 impressed is assumed, 
then 

100 



i^i = 



= 6.62 



4X3.7764 

The next step is to choose between 6 and 7 for Kij, As it would 
require a higher flux density in the gap if Ki should equal 6, 
take Ki as being 7. Practically, this choice means that the 
motor will have a high power factor but possibly slightly lower 
efficiency. The preliminary value of gap density will be decreased 
from 25,000 to the necessary value to give a back-e.m.f. of 100 
volts: 

/g = 25000X^ = 23650 lines per sq. in. ^ 

Main Winding. Since Ki is to be 7, the number of conduc- 
tors per slot will be 14 in the slots shown with 2 conductors and 
7 in those shown with 1 conductor. The number of turns per 
pole will be 70, the number of conductors per pole 140, and the 
total number of conductors 560. From Fig. 29 the slot area will 
be 0.75X0.203, or 0.1522 square inch. Allow for a 0.015-ineh 



75 



76 DESIGN OF SMALL MOTORS 

slot lining and a slot wedge at the top 0.04 inch thick, leaving 
0.1122 square inch. Assume that 0.5 of this will be copper sec- 
tion, or 0.5X0.1122, which equals 0.0561 square inch. As the 
starting winding will occupy some of the slots which contain the 
full number of primary conductors, assign 0.85 of the total copper 
section to the main winding. The size of conductor will thus be 
0.0561X0.85^ ^^ Q QQ3^ ^^^^^^ j^^j^ ^t^^ ^4 -^ 0.003225 square 

inch, and double cotton-covered No. 14 will, therefore, be used 
for the main winding. 

The length of mean conductor will be about that of the 
conductor in the slot between teeth 2 and S. The end connection 
will extend over nine teeth, a distance which, when laid off at 
the center of the slots in Fig. 29, is 3.25 inches. The active 
length of the conductor will be 2.5 inches, but \ inch must be 
allowed at each end, making the total length of mean conductor 
3.25-1-0.75+2.5+0.75, or 7.25 inches, or 0.604 foot. The total 
length of wire in the main winding will be 0.604X560, or 338 
feet. The hot resistance of No. 14 is 0.002823 ohm, making the 
total resistance of the main winding 338X0.002823, or 0.955 ohm. 
Before proceeding further, the magnetizing current and the power 
factor should be estimated. 

Magnetizing Current. The total ampere turns will be com- 
puted in the same way as for a d.c. machine, using Fig. 22 to obtain 
the ampere turns per inch. The primary contraction factor will 

TF/+2A 
be , where Wt is the tooth width; A, the air-gap length; 

and V, the slot pitch. Substituting 0.2885 for Wt, 0.012 for A, 

J AH10C: p .X. ^ 1 u 0.2885+0.024 0.3125 

and 0.4135 tor 75, the lormula becomes , or , 

^ 0.4135 0.4135 

which equals 0.756. The secondary contraction factor will be 

0.478+0.024 0.502 ,. , , „^_. 

0.538 - '^0538' '""^'^^ '^^'^' ^•^^^'- 

The effective part of the primary tooth pitch will be 0.756 X 

0.9335X0.4135, or 0.292 inch. The various values of flux pre- 

viously computed will be reduced in the ratio z— 777^7: on account 

2o000 

of the reduced flux density in the air gap. 



76 



DESIGN OF SMALL MOTORS 



77 



Part Flux 


Section 


Flux 
Density 


Ampere 

Turns 

per Inch 


Length 
(in.) 


Ampere 

Turns 

per Pole 


Primary core 185600- 


-2 0.92X2.5 


40300 


4.0 


1.5 


6.00 


One tooth 
(maximum density) 24350 


0.15X2.5 


65000 


10.0 


0.75 


7.50 


Air gap (one tooth, 

maximum density) 24350 


0.292X2.5 


33300 


10440.0 


0.012 


125.10 


Secondary core* 175000- 


-2 0.8X2.5 


43800 


4.5 


3.0 


13.50 


Total 


152.10 



The 152.10 ampere turns per pole obtained in the tabulation 
must be multiplied by 1.9 to allow for the part of the excitation 
supplied by the secondary winding, making 1.9X152, or 288.5 

288.5, 



ampere turns maximum value. The effective value will be 



V2 



or 204 ampere turns. As the number of turns per pole on the main 

winding will be 70, the effective value of the magnetizing current 

204 
will be -— -, or 2.915 amperes. In order to judge whether this is 

satisfactory, assume that the final design should have an efficiency 

of 70 per cent and a power factor of 0.75. The total current 

746 X - 
will then be — — — ^ „ "^^ ^^ , or 6.46 amperes. The power com- 
110X0.7X0.75 ^ ^ 

ponent will be 0.75X6.46, or 4.85 amperes. The reactive component 
will be 'v/6. 46^—4.852, or 4.29 amperes. The ratio of the magnetiz- 
ing current to the estimated total reactive component of current 
2.915 



will be 



4.29 



or 0. 



Since this ratio should be between 0.85 



for very small motors and 0.6 for large motors, the ratio obtained 
is satisfactory. Another check on the general proportion of the 
design is the number of in-phase component ampere conductors 
per inch of periphery, which should be less than 250 for motors 



of 1 hp. or under. 



In this case it will be -f---r , or 136.7. The 

6.312X7r 

design may now be proceeded with. 

Starting Winding. The starting winding should have from 
45 to 65 per cent of the conductors of the main winding. Taking 
55 per cent, the number of conductors will be 560X0.55, or 308. 
In order to leave as much room as possible for the main winding, 



The ampere turns required by the secondary teeth arc negligible in this case. 



77 



78 



DESIGN OF SMALL MOTORS 



the starting winding will be arranged as shown in Fig. 31, where 
those slots containing the larger number of conductors fit into 
the slots which contain only one-half the full number of main 
conductors. Forty-eight conductors are shown in Fig. 3L These 
are to be multipHed by a factor K2, so that 



and 



48X2 = 308 
48 



If K2 is taken as 6, there will be 6X48, or 288 conductors in 
the starting winding. The resistance of the starting winding must 

TYPE SA rom SI. FRAME. 165. STAPT/N6 Wm/N6 




Fig. 31. 



Starting Winding for Western Electric Type SA, Form SI, Frame 165, 
Single-Phase Induction Motor 



4 



be high enough to secure a current which is as near in phase with 
the line voltage as possible but low enough to give a substantial 
A'alue of current. A good compromise is four times the normal 
full-load current, which was estimated as 6.46 amperes. The 

110 



impedance will thus be 



:, or 4.265 ohms, of which the 



4X6.46^ 

resistance will be about 4 ohms. From Fig. 29 the length of 
mean conductor will be 3.25+1.25+2.5 + 1.25, or 8.25 inches, which 
equals 0.688 foot. This is slightly greater than the length of mean 
conductor in the main winding, owing to the fact that the start- 
ing winding is to be wound on over the main winding. The 
total length of wire will be 0.683X288, or 197 feet. The resist- 



78 



DESIGN OF SMALL MOTORS 79 

245 
ance per foot must therefore be — -, or 0.0203 ohm. No. 23 has 

a cold resistance of 0.02036 ohm per foot and its cross-section is 
0.0004002 square inch. In the slots which are to contain con- 
ductors of both the main and the starting windings, the number 
of conductors in the starting winding will be 6, giving a copper 
section of 6X0.0004002, or 0.0024 square inch. The number of 
main conductors in this same slot will be 14 of No. 14, having 
a section of 14X0.003225, or 0.0451 square inch. The total 
copper section in the slot will thus be 0.0451+0.0024, or 0.0475 
square inch out of a total winding space of 0.1122 square inch, 
which will therefore afford sufficient room. 

Secondary Winding. The secondary member will be wound 
with a squirrel-cage winding completely filling the slots, which 
are to be circular and 0.3 inch in diameter. In order to obtain 
about the same ctirrent density, the end ring should have a sec- 
tion equal to about one-third of the total section of the bars per 

pole, or |X-^X-j-X0.302, ^j, o.218 square inch. The end ring 

is employed in this frame as a mechanical support for the end 
shields and will be 1 inch wide. Its thickness will therefore be 
0.218 inch. The volume of copper in the secondary winding will 

be: 37 bars, 37X^X0.302X2.5, or 6.54 cubic inches; 2 end rings, 

2X0.218x7.34X'7r, or 10.05 cubic inches. 

Losses. The volume of iron in the tooth necks will be 
0.15X0.75X2.5X48, or 13.5 cubic inches. The flux density in 
the teeth was found to be 65,000 lines per square inch. Using 
Hobart's formula for 0.014-inch laminations at 60 cycles 

Watts per cu. in. =4. if — ^ 

in which f^m is the maximum flux density, the total teeth iron 
loss will be 4.1 (0.422) X 13.5, or 23.35 watts. The volume of 
active iron in the primary core will be 0.92 X2.5X 3.68 Xtt, or 
26.6 cubic inches. The core flux density was found to be 40,300. 
The core iron loss will be 4.1(0.162) X26. 6, or 17.67 watts. 
The total iron losses will therefore be 23.35 + 17.67, or 41.02 watts. 



79 



80 DESIGN OF SMALL MOTORS 

The bearings will be 0.75 inch in diameter and 3.5 inches 
long. Assume a speed of 1700 r.p.m., which will make the rub- 

bing velocity in the bearing ' X1700, or 334 feet per minute. 

The bearing loss will be 0.8 dl ( ^7^) watts, where d and I are the 

bearing diameter and length, respectively, and v is the rubbing 

velocity. For two bearings the loss will be 2X0.8X0.75X3.5X 

3.44% or 26.8 watts. 

The brushes which bear on the two slip rings, Fig. 8, will 

have a section sufficient to carry 7 amperes at 40 amperes per 

7 
square inch, that is, — , or 0.175 square in^ch. They may be 

made 0.6 inch X 0.3 inch. Assuming a pressure of 2 pounds per 

square inch and a coefficient of friction of 0.3, the friction loss 

47r 
with rings 4 inches in diameter will be 2X0.18X2X0.3X — X1700, 

384 

or 384 foot pounds per minute, which is equivalent to -; X 746, 

ooUUU 

or 8.67 watts. Allowing a brush drop of 2 volts and the total cur- 
rent being 6.46 amperes, the brush contact loss will be 2X6.46, 
or 12.92 watts. Allow 3 watts for windage loss. 

The PR loss in the secondary winding may be estimated by 
finding the component of current in the primary corresponding 
to the total mechanical power output transmitted across the air 
gap, which output is as follows: 

Power delivered | X746 373.0 watts 

Bearing loss 26.8 watts 

Brush friction 8.67 watts 

Windage 3.0 watts 

Total 411.47 watts 

The back-e.m.f. of the primary has been taken as 100 volts, so 
that the desired component of primary current will be '- — , 

or 4.115 amperes. The total effective value of the primary mag- 
netizing current was found to be 2.915 amperes, out of which 

0.9 

-^2.915, or 1.381 amperes, will be transformed into the secondary 

m a single-phase motor, making the total transformed current 



80 



4 






DESIGN OF SMALL MOTORS 81 

V^4.1152+L382, or 4.34 amperes. Since there are to be 140 
conductors per pole, this will give 4.34X140, or 607.5 ampere 
conductors per pole, which will be opposed by a like number of 
ampere conductors in the secondary. The current in the 9i 
secondary bars per pole will be distributed in approximately 
sine-wave proportion, so that the maximum ampere conductors 

TT 1 

per bar will be found by plotting to be -—Xr—- X 607.5, or 103.3 

1 0'^ "^ 
amperes per bar, which will give a current density of - — — — , 

^XO.32 

or 1462 amperes per square inch. From Fig. 18 this corresponds 

to 1.7 watts per cubic inch, making the watts loss in the bars 

1.7X6.54, or 11.1 watts. The current in the end rings will be 

607.5 ampere conductors -r-2, or 303.75 amperes; the amperes 

Q ri Q '^ r' 
per square inch will be ' ' , or 1395, which corresponds to 

0.218 

1.55 watts per cubic inch, making the watts loss in the rings 

1.55X10.05, or 15.57 watts. The total secondary loss will thus 

be 11.1 + 15.57, or 26.67 watts. 

The estimate of 1700 r.p.m. for the full-load speed may now 

be checked up. The in-phase component of secondary current is 

to be 4.115 amperes (equivalent primary value) and the loss due 

/4.115\^ 
to this component will be ( -j-^r ) X 26.67, or 24 watts. Since the 



4.34 

24 
transformed input is to be 411.5 watts, the sHp will be -^ — X 

411.5 

1800, or 104.8 r.p.m. The full-load speed will therefore be 1695 

r.p.m., which is satisfactory. 

The primary copper loss can only be estimated until the 
efficiency and the power factor, and hence the full-load current, 
have been determined. Losing the estimated value of 6.46 amperes, 
the primary PR loss will be 6.46^X0.955, or 39.8 watts. 

The total losses will be as follows: 

Primary PR loss 39.80 watts 

Secondary PR loss 26. G7 watts 

Iron losses '. 41 .02 watts 

Bearings and windage 29.80 watts 

Brush contact loss 12.92 watts 

Brush friction 8.67 watts 

Total 158.88 watts 



81 



82 DESIGN OF SMALL MOTORS 

Summary. The output of the motor is to be iX746, or 373 

373 373 

watts. The efficiency will thus be gyg, ^5ggg > ^^ 531^' ^^ ^^ 

per cent, which happens to agree exactly with the original estimate. 
Since the voltage is to be 110, the primary power component of 
current will be 531. 88 -i- 110, or 4.84 amperes. It is not possible 
to obtain the reactive component of primary current without 
computing the impedance of the primary and the secondary wind- 
ings. It may be assumed, however, without serious error that the 
magnetizing current will be 70 per cent of the total reactive com- 
ponent. This percentage is greater the greater the value of mean 
gap flux density. Thus the reactive component of the primary cur- 
rent will be ' , , or 4.17 amperes. The total current will there- 
0.70 



fore be V4.842 + 4.17^ or 6.39 amperes, which is close to the 
original estimate, so that it will not be worth while to revise the 
primary PR loss and the efficiency. The corrected value of power 

4 84 
factor will be -^, or 0.757, which is close to the estimate of 
6.39 

0.75. The primary IR drop will be 6.39X0.955, or 6.1 volts. 
The impedance drop will probably be less than three times this 
and will be added to the assumed value of 100 volts back-e.m.f. 
at an angle, so that 100 volts will not be far from correct. In 
general, a high value of ampere conductors per inch of periphery 
will give a greater impedance. The practical effect of a small 
discrepancy between assumed and actual back-e.m.f. will be to 
vary the flux density slightly and will merely result in varying 
the power factor somewhat. The computation of the starting 
torque is beyond the scope of this treatment. 

The heating of the motor is determined as follows: Refer- 
ring to Fig. 28, the radiating surface will be irD(—-\-L\ or 7r8.75 

8 75 \ 

-- — h8 ), or 341 square inches. The peripheral velocity will 



be -4^X1700, or 2805 feet per minute, which will give 0.012 
wiitt per square inch for TC;. rise. With 158.88. watts loss, the 



82 



DESIGN OF SMALL MOTORS 83 

158 88 
rise, if the motor were enclosed, would be ' , or 

38.7° C. As this design is to be well ventilated and to have a 
small punched fan attached, it is certain that the rise will be 
satisfactory. 

The shop information on the ^-hp., 110- volt, single-phase 
induction motor will be as follows: 

Mechanical Data Frame No. 165, type SA. 

Electrical Data 

Main Winding : As per Fig. 30. 560 conductors of No. 14 double 

cotton-covered wire. 

Starting Winding : As per Fig. 3 1 . 288 conductors of No. 23 double 

cotton-covered wire. Wind over main wind- 
ing and connect in parallel with same through 
centrifugal cutout to slip rings. 

Secondary Winding : Standard end rings, 1X0.218 section. 

All other details standard with No. 165 frame. 

|=HORSEPOWER 25=CYCLE MOTOR 

See Case VII under Changing Characteristics of Motor by 
Rewinding. 

DESIGN OF INDUCTION=MOTOR FRAMES 

A good example of proper induction-motor frame design is 
shown in Fig. 24. In many cases, however, the squirrel-cage or 
secondary member is made the rotor. 

As in the case of d.c.-motor frame design, the fundamental 
problem is to secure a good magnetic circuit and at the same time 
provide sufficient space for the windings. An instructive exercise 
is to vary the slot width, depth, etc., and compute the magnetiz- 
ing current by the method previously given. It will be found 
that the proportions of Figs. 24 and 29 are ideal when both wind- 
ing space and magnetic requirements are considered. 

The air gap in induction motors is an important detail of 
design. Magnetically, the smaller the gap the better, but for 
mechanical reasons it is necessary to employ a reasonable clear- 
ance. Hobart recommends the formula 0.015v^kw. inches, where 
kw. is the power input in kilowatts. 



8S 



DESIGN OF SMALL MOTORS 

PART II 



CHANGING CHARACTERISTICS OF MOTOR 
BY REWINDING 

DIRECT=CURRENT MOTORS 

G)nditions to Be Met. A practical problem which is often 
met with is that of changing the voltage, speed, etc., of an existing 
motor which may or may not be in good condition and may even 
be without a name plate. It is possible to take such a motor and 
by making certain measurements and tests obtain sufficient infor- 
mation to compute new windings which will make the motor 
meet the new requirements, provided of course these are within 
the capabilities of the size of frame considered. 

) Direct-current motors will first be considered. Two methods 
of procedure will be outlined; first, in case the windings of the 
motor have been damaged and the name plate is lost so that no 
information is available; second, in case the motor has windings 
in good condition and has a name plate giving the usual infor- 
mation. 

CASE I 

Necessary Data. The data which the designer must know 
is as follows: 

Given: Direct-current motor complete, except windings burned out and no 

name plate. 
Required: Rewind to run on 32 volts at 2000 r.p.m. and develop as much 

power as consistent with good efficiency. 

Armature. The armature should first be removed and the 
following measurements taken and tabulated: 

Armature 

Armature diameter 3.0 inches 

Number of slots 24.0 

Slot depth below surface 0.5 inch 

Average slot width 0.25 inch 



86 



DESIGN OF SMALL MOTORS 



Armature 

Width of tooth at neck 0.10 inch 

Width of slot opening 0.08 inch 

Useful slot depth 0.45 inch 

Armature length 2.0 inches 

Air gap 0.03 inch 

Commutator bars 24.0 

Commutator diameter 2.0 inches 

Field 

Number of poles 2.0 

Pole arc 3.2 inches 

Pole-face area 6.4 square inches 

Cross-section of pole core -2.5 square inches 

Winding space available for field coils. . 0.5 square inch 

Cross-section of yoke 1.7 square inches 

Cross-section of armature 1.7 square inches 

General information as to the proper mean pole-face flux 
density is given in Fig. 32. This information is approximate, 



§ 40 



rWX DENSITY FOR SMALL D.C MOTORS 



50 



20 



10 















_ 


































- 






































^ 






'-1 






























^ 




■* 




































^ 




^ 








































^ 


<** 










































/" 














































/ 














































/ 
















































































































































A 


IQTF: Th 


p / 


ih 


IW 


? inforn 


nnfinn nnn/ip<i to 


— 


- 




































III 


J IK. 


/I J 






1 n 


cu 


1 a 


/ ^ 


lU^ 


JCl 




1JU 


lO. 



















































































































































12 3 4-56 

DIAMETER OF ARMATURE IN INCHES 

Fig. 32. Flux Densities for Small Direct-Current Motors 

but it is possible to employ densities within 10 per cent of those 
given without encountering serious trouble. Therefore, it will be 
advisable to assume a mean pole-face density of /^ to equal 32,000, 
corresponding to an armature diameter of 3 inches. The flux per 
pole will be 6.4X32000, or 204800 lines. Assume a back-e.m.f. 
of 28 volts out of the impressed 32, leaving 4 volts for brush drop 
and armature IR drop. The number of inductors can be found 
from the voltage equation as follows: 



86 



DESIGN OF SMALL MOTORS 87 

60 c 

9000 2 
28= iVx204800X^X^XlO-8 

iV = 412 inductors 

412 
which equals — — , or 17.2 inductors per slot. It will be wise to 

use 18 conductors per slot, or 432 inductors, and reduce the pole- 

17 2 

face flux density to 32000 X—^, or 30550 lines per square inch. 

18 

The flux per pole will be 6.4X30550, or 195600 lines. The useful 

slot area will be 0.45X0.25, or 0.1125 square inch. Employ a 

slot lining 0.015 inch thick and leave room for a slot wedge 0.06 

inch thick. The net area available for winding space will thus 

be reduced to 0.0765 square inch. Assume that one-half of this 

will be copper cross-section, making the copper section per slot 

0.03825 square inch. The inductor section will thus be 0.03825-4- 

18, or 0.002125 square inch. No. 16 is 0.002028 square inch, 

which is sufficiently close. The armature winding will therefore 

consist of 18 inductors per slot of No. 16 double cotton-covered wire. 

Use will be made of a simplex singly re-entrant drum winding of 

10/24 pitch, or in other words a coil will enclose 10 teeth out of 24. 

The length of mean inductor will be the sum of the armature 

length, the end arc, and the corner allowance, which equals 

24-3+1.5, or 6.5 inches. The total length of armature wire will 

6 5 
thus be 432 Xt^, or 234 feet. The hot resistance per foot of 

No. 16 is 0.004489 ohm. The resistance of the armature will thus 
be iX 0.004489X234, or 0.2625 ohm. From Fig. 33 it is evident 
that a reasonable value for the number of ampere conductors per 
inch of armature periphery will be 185. The armature current 

in each conductor will thus be — — — — , or 4.03 amperes. The 

total armature current will be 8.06 amperes. The armature IR 

drop will be 2.115 volts, and the PR loss 17 watts. The ampere 

, .„ , 432X4.03 .^- 

turns per pole will be — - — - — , or 435. 

2X^ 



87 



88 



DESIGN OF SMALL MOTORS 



Fieid. The number of ampere turns required for the field 
may be determined by the usual tabulation of ampere-turn require- 
ments for each part, as follows: 



Part 
Yoke 
Pole core 
Air gap 
Teeth 
Armature core 



Flux* 
190000^2 
220000 
195600 
185000 c 
180000 H- 2 







Ampere 




Ampere 




Flux 


Turns 




Turns 


Section 


Density per Inch Length 


per pole 


1.7 


55900 


7 


2.5 


18 


2.5 


88000 


30 


0.5 


15 




32000 


10020 


0.03 


301 


.1 inch X 2 inches 102700 


80 


0.45 


36 


1.7 


53000 
Total 


6 


2.0 


12 
. 382 



AMPERE- CONDUCTORS PIU IN. FOR SMALL D.C.MOTORS 



6b0 








~1 




















— 






















ZOO 




























































































--^ 




■ 


^250 


































































































|200 




























^ 


x*' 










































y 


y 
























9- 






















y 












































/ 


/ 






























^100 
















/ 














































/ 




































50 













































































































































L 







\ 2 Z 4r 5 6 

DIAMETER OF ARMATURE IN INCHES 

Fig. 33. Ampere-Conductors per Inch for Small Direct-Current Motors 

The flux density in the air gap is determined from the mean 
pole-face density by computing the slot contraction factor in the 
way given under the previous illustrative designs. 

The field will require 382 ampere turns at no load. The full- 
load ampere turns will be approximately 1.25X382, or 478. Using 
the estimated value of current obtained from Fig. 33, it was found 
that the number of armarture ampere turns per pole was 432. 
The ratio of field ampere turns to armature ampere turns will 

thus be — — , or 1.105. As a general rule this ratio should be 

about 1.25. The indications are, therefore, that the first estimate 
of 8.06 amperes armature current may have to be reduced. This 
will be decided, however, after determining the heating. 

* The values of flux have been altered in the various parts to allow for leakage. 



88 



f 



n DESIGN OF SMALL MOTORS 80 

\^m- The winding space available for the field coils will be 0.5 

square inch. Assume that after allowing for insulation the total 

copper section will be 0.22 square inch. This will give a current 

478 
insity in the field of ^r^^, or 2170 amperes per square inch. 

This density should be between 2000 and 3000 per square inch 
for motors under J hp. and between 1500 and 2500 for motors 
between | hp. and 1 hp. From Fig. 18 the watts per cubic inch 
will be 3.65, corresponding to 2170 amperes per square inch. As 
the length of mean turn is to be 10.5 inches, the watts loss in the 
two field coils will be 2X3.65X0.22X10.5, or 16.85 watts. Since 
the voltage is to be 32, if a shunt field is used, the current in 

the field circuit will be , or 0.5275 ampere. At 2170 amperes 

5275 

fr square inch the cross-section of the field wire will be ' , 
0.000243 square inch. No. 25 is 0.0002517 square inch. There- 
fore the field will be wound full with No. 25. Since the total 
^.. . , 1 ..^ . . XT .. .1. . 0.22 



)per section is to be 0.22 square inch, No. 25 will give 



1 

IR) 



0.0002517' 
or 875 turns per coil. The total length of wire for two coils will 

'—, or 1530 feet. The resistance of No. 25 is 

12 

'3618 ohm per foot, making the resistance of the field circuit 

1530X0.03618, or 55.25 ohms. The current will be — — , or 

50.25 

0.58 ampere, and the ampere turns per pole developed will be 
0.58X875, or 508, which is somewhat more than originally com- 
puted but which will be satisfactory, merely raising the flux den- 
sity a trifle. The ratio of field to armature ampere turns per 

508 
pole will now be -— , or 1.173, which is nearly 1.25 and may be 

considered satisfactory. 

Rating. The windings have now been computed, and it 
merely remains to determine how much load the motor will carry. 
This involves determination of the losses and the resulting tem- 
perature rise, which may be estimated approximately from Fig. 28. 
The procedure is given in detail in the previous illustrative designs 



90 DESIGN OF SMALL MOTORS 

and may be worked out if desired. For illustrative purposes, 
assume that the estimated value of 8.06 amperes armature current 
will give losses which result in a moderate temperature rise and 
that it may therefore be taken as the permissible full-load arma- 
ture current. The field current will be 0.58 ampere, making the 
total current 8.64 amperes and the total power input 277 watts. 
By subtracting the losses found by the methods employed in the 
illustrative designs the useful output can be obtained, and this \ 
divided by 746 watts gives the horsepower of the motor. 

The efficiency of a motor of this size would probably be 

about 65 per cent, so that its output would be — -^ — , which 

equals 0.241 hp., or approximately | hp. The windings for thisi 

J-hp. motor will be as follows: 

Field: 875 turns per coil of No. 25 single cotton-covered wire. 

Armature: 18 inductors per slot of No. 16 double cotton-covered wire. 

CASE II 

Necessary Data. The data for the motor to be rewound is 
as follows: 

Given: Motor in good condition with the following information on the 

name plate: direct current motor, ^ horsepower, 32 volts, 8 amperes, 

2000 r.p.m. 
Required: Rewind, making an automobile charging generator suitable for a 

3-cell lead battery. Cut-in speed 1500 r.p.m.; maximum speed 

5000 r.p.m. Power output as large as possible. 

For convenience it is assumed that the motor which has 
just been worked out under Case I is the given machine. Reference 
should therefore be made to Case I for the various dimensions and 
other data quoted herein. The motor will be rewound and 
equipped with a series and a shunt field. The direction of winding 
of the series field will be such that when the generator is 
charging the battery the series field opposes the shunt field, so 
that as the speed is increased the increased current will reduce 
the flux and thus limit the rise in current to a reasonable value. 

In a design of this sort it will be necessary to employ a lower 
value of mean pole-face flux density than is given in Fig. 32 for 
an armature 3 inches in diameter. This is due to the fact that 
field winding space must be divided between the series and the 



90 



DESIGN OF SMALL MOTORS 91 

shunt windings, and as these windings oppose each other the 
resultant ampere turns are less for the same current density. The 
limit of current density is the heating. The field coils will there- 
fore be designed first. 

Field. The operating condition which will develop the great- 
est heat in the field coils will occur at maximum speed with a 
fully charged battery. The voltage will then be about 2.45 volts 
per cell, or 3X2.45, which equals 7.35 volts at the battery, or 
about 7.6 volts at the generator, allowing a J-volt lead drop. 
As the generator will be running at maximum speed, which is 

, or 3.33 times the cut-in speed, the flux will be in the neigh- 

1500 

borhood of — — of that at cut-in speed, or, in other words, the 

o.oo 

resultant ampere turns will be — — of the ampere turns developed 

o.oo 

by the shunt field alone, since at cut-in speed the current in the 
series coil will be negligible. Let Fi and F2 be the shunt- and the 
series-field ampere turns, respectively, at maximum speed, and 
the following relation will be true: 

Fi 

or 

F 2=0.7 Fi 

As a matter of fact, the armature reaction at high speed helps 
to demagnetize the field, so that this value of F2 is somewhat 
higher than necessary. It is a good plan, however, to design the 
series field liberally, and therefore the proportion F 2 = 0.7 Fi will 
be retained. 

Owing to the fact that the maximum speed will probably 
not be maintained indefinitely, a somewhat higher current density 
may be employed in the series field, and therefore 3200 amperes 
per square inch will be allowed for it and 2500 amperes per square 
inch for the shunt field. Let Si and ^2 be the total copper cross- 
sections of the shunt and the series windings; then 

320()8_2^^^ 
25OOS1 Fi ' 

91 



92 DESIGN OF SMALL MOTORS 

or 6^2 = 0.7x|^5i = 0.547Si 

But 

^^-(-52 = 0.22 square inch copper section of old winding 

This section may be obtained by counting the number of turns 
and multiplying by the section of the wire. Thus 

5i+0.547Si = 0.22 sq. in. 

22 
•51 = 7^ = 0.1422 sq. in. 
L54/ 

& = 0.547X0. 1422 = 0.0778 sq. in. 



therefore 



^1 = 2500X0.1422 = 355 amp. turns 
^2 = 3200 X 0.0778 = 249 amp. turns 



Shunt Winding. Place the shunt winding next to the pole 
cores. Its length of mean turn will thus be 9 inches. The vol- 
ume of copper in both shunt-field coils will be 2X0.1422X9, or 
2.56 cubic inches. From Fig. 18, at 2500 amperes per square 
inch the watts per cubic inch will be 4.75 watts. The power 
loss in the shunt field will thus be 2.56X4.75, or 12.16 watts. 
The voltage was previously computed as 7.6, making the shunt- 
field current ^' , or 1.6 amperes. The section of the shunt-field 

wire will therefore be ^— tt:, or 0.00064 square inch. No. 21 

zoUU 

is 0.0006363 square inch. Therefore use No. 21 single cotton- 
covered wire for the shunt-field coils. The number of turns will 

^^^2 
be .vw>or'> > ^^ ^^-^^ turns per coil. The total length of wire 
O.OOOboOo 

9 
will be 2X223X— , or 334.5 feet. Since No. 21 has a resistance 

of 0.01431 ohm per foot, the resistance will be 0.01431X334.5, 
or 4.79 ohms. Since the voltage is to be 7.6, the revised value 

of the shunt-field current will be — ^, or 1.587 amperes, giving 

Fi = 1.587X223 = 354 amp. turns 
which checks with the value desired. 



92 



DESIGN OF SMALL MOTORS 93 

It is now possible to judge how much below normal it will 

be necessary to make the mean pole-face density. Having counted 

the number of turns of the old field winding and measured the 

resistance, it is found that with the old voltage of 32 the former 

field current was 0.58 ampere and the former field ampere turns 

354 
508. The ratio of new to old ampere turns will be — — , or 0.698. 

508 

Owing to partial saturation with the old winding, it is reasonable 

to expect the new flux to be 75 per cent of the old. 

Series Winding. Place the series winding over the shunt, 
thus making the length of mean turn 13 inches, as may readily 
be determined by laying out the dimensions of the core and the 
shunt coil. The volume of copper for both series coils will be 
2X13X0.0778, or 2.022 cubic inches. At 3200 amperes per square 
inch the watts per cubic inch are 7.8, Fig. 18, making the series 
field watts 2.022x7.8, or 15.78. The value of current output of 
the generator and the size of wire for the series field will be left 
for a later decision. It may be noted, however, that the total field 
wattage will be 12.16 shunt+ 15.78 series, or 27.94 watts. These 
computations could be checked experimentally for temperature rise 
by applying sufficient voltage to the old field coil (which has not 
yet been unwound) to make it dissipate 14 watts. If the tem- 
perature reached by the coil within one hour is not too hot for 
the hand to bear, the proposed field windings will be satisfactory. 
If the temperature rise is too high, it will be necessary to employ 
lower current densities in the field coils and to revise the wind- 
ings accordingly. 

Armature Winding. The computation of the armature wind- 
ing is preferably based on the cut-in speed when the current will 
be small. Assuming the battery to be fully charged, the generator 
voltage will be 7.35 volts. The J-volt drop in the leads is here 
omitted since the current is to be small. Metallized carbon brushes 
such as the Le Carbone Company's grade Metal No. 2 will be 
employed. The brush drop will be about 0.3 volts, making the 
internally generated e.m.f. 7.65 volts. 

It was found after computing the shunt-field winding that 
the new flux may be expected to be 75 per cent of its former 
value. As this is an approximate value, however, only 65 per 



93 



94 DESIGN OF SMALL MOTORS 

cent will be counted on, and it will thus be ensured that the 
generator will be producing sufficient voltage when cut in at 1500 
r.p.m. Assuming that the old armature winding was correctly 
designed, the generator winding may be computed by proportion. 
The number of inductors per slot will be 18 old winding X 
7.65 volts generated e.m.f. 1 2000 r.p.m. old speed ^ ^^^^ 

28 volts back-e.m.f.* 0.65 1500 r.p.m. new speed' 
inductors per slot. Ten inductors per slot, or 240 inductors total, 
will therefore be used. The total copper section of the old arma- 
ture winding was 0.03825 square inch per slot. Using the same 
total copper cross-section, the size of the new armature inductor 

will be ^J^^^, or 0.003825 square inch. No. 13 is 0.004067 

square inch, which is a trifle larger, but the copper section can be 

safely increased a little as the space factor will be better with the 

larger size of wire. 

The armature winding will thus be 10 inductors per slot of 

No. 13 double cotton-covered wire. A simplex singly re-entrant 

10/24-pitch winding will be used. The length of mean inductor 

will be 6.5 inches, the same as in the old winding. The total 

6 5 
length of armature wire will be 240 X-^, or 130 feet. The resist- 

ance per foot of No. 13 is 0.002239, making the armature resist- 
ance 1X130X0.002239, or 0.07275 ohm. The number of ampere 
inductors per inch of periphery in the old winding can be obtained 
from the current on the name plate and the number of inductors, 

T . 432 inductors X 4 amperes per inductor ^^„ r 

and IS — -. — . , , or 183. In view 

6T mches periphery 

of the fact that a weaker field is being used than in the old design, 

a lower value for the ampere inductors must be employed so as 

to preserve a proper ratio between field ampere turns and armature 

ampere turns. Reducing the 183 ampere inductors in the same pro- A\ 

portion as the field, the result will be 0.65X183, or 119 ampere 

inductors per inch of periphery. Even this value, however, will 

be too high at the higher generator speeds, since the action of 

the differential field reduces the flux. Therefore take a value of 



* This value is assumed after allowing for the brush drop and the armature IR drop out of 
the 32 volts suppUed. 



94 



DESIGN OF SMALL MOTORS 95 

75 ampere inductors per inch of periphery, which will give good 

119 
commutation up to a speed of ISOOX-^irr, or 2380 r.p.m. Beyond 

^lis, speed commutation will be somewhat poorer, but this will 
not be very serious in a generator of such low voltage. 

f The inductor current in the new winding will be , or 

^■±u 

2.95 amperes. The current density in the armature inductors will 

2 95 
be '- 3, or 726 amperes per square inch. This is a very low 

density, and a smaller size of wire could be used if there were 
anything to be gained. 
■ Rating. The generator will be rated at 2X2.95, or 5.9 
amperes. It is now possible to finish the design of the series 
field. The current density is to be 3200 amperes per square inch 

at maximum speed. This will give a conductor section of 5^^, 

or 0.001843. No. 16 is 0.002028 square inch. As the series 

winding is to be wound on over the shunt, the original allowance 

for coil section may be exceeded somewhat and No. 16 used. 

Since there will be 249 ampere turns, the number of turns per 

249 
coil will be — -, or 42.2 turns, which will be considered 42 turns 
5.9 

per coil. The total length of wire for the two coils will be 2X42X 

j|, or 91 feet. The resistance for No. 16 wire will be 0.004489 

X91, or 0.4085 ohm. The IR drop at maximum speed will be 
5.9X0.4085, or 2.41 volts. Connect the series winding inside the 
shunt winding so that the latter will be across the generator 
terminals. 

The windings for the generator are now completely deter- 
mined as follows: 

Shunt field: 223 turns per coil of No. 21 single cotton-covered 
Series field: 42 turns per coil of No. 16 double cotton-covered 
Armature: 10 inductors per slot of No. 13 double cotton-covered 

If desired, the efficiency may be computed by the methods 
worked out in the preceding illustrative designs. As both the 
flux and the current densities are very low in the armature, a low 



95 



96 DESIGN OF SMALL MOTORS 

temperature rise may be expected for the generator at medium 
speeds. The generator may be rated as 6 amperes maximum 
charging rate. As in all generators of this class, the charging rate 
increases with increased speed. At a cut-in speed of 1500 r.p.m. 
it would be very small, probably between and 2 amperes. At 
2000 r.p.m. it would be from 3 to 4 amperes and would increase 
up to 6 at 5000 r.p.m. The generator would thus be suitable 
for use with a three-cell lead battery of from 30 to 60 ampere 
hours capacity. 

CASE III— CHANGE OF VOLTAGE 

Necessary Data. The motor to be rewound is as follows: 

Given: Direct-current motor, | hp,, 55 volts, 1700 r.p.m. Shunt wound. 

In good condition. 
Required: Rewind so as to make a 110-volt d.c. motor of f hp. at 1700 r.p.m. 

All that is necessary in this case is to rewind the field so as 
to give the same ampere turns and the armature so as to make 
the back-e.m.f. the same percentage of the impressed e.m.f. The 
data required may be obtained by counting the turns of the pres- 
ent windings and increasing them by the necessary ratios. Hi 

Field. Since the ampere turns are to be the same and the 
new field coils will have the same volume, the current density 
in the field we will be the same, and the power loss will thus be 

the same. Since this power is suppHed at -r^, or 2, times the 

55 

voltage, the new field current will be one-half the old and the 

section of field we will be one-half the old. In a B.&S. gage 

wire the section is reduced one-half by choosing a gage of wire 

three numbers higher. The number of field turns will be doubled 

and the field resistance will be four times its former value. 

Armature. Since the field ampere turns are to be the same, 

the fiux will be the same. The same speed is also specified. The 

armature inductors will therefore be doubled, as will be evident 

by doubhng e in the voltage equation 

60 C 

Since the slot section is the same, each inductor will be one-half 
the section of the former inductor, which means the use of a gage 
of wire three numbers higher. 



96 



DESIGN OF SMALL MOTORS 97 

Caution. In the above treatment it has been assumed that 
the space factor, or ratio of copper section to coil section, has not 
been noticeably changed. No appreciable error will be introduced 
because of this assumption except when changing from a very 
large size to a very small size or when the thickness of insulation 
is not changed in the same proportion as the size of wire. In 
such cases the number of armature inductors may be computed 
by the method given, but their section will be greater or less than 
the section so computed according to whether the space factor 
is improved or not, respectively. Figs. 19 and 20. The field 
ampere turns should be kept the same regardless of the number 
of turns or the size of wire. In the case of a poorer space factor, 
this would mean a greater power loss in the field circuit and a 
higher current density in the windings. 

It is obvious that the voltage of a motor may be increased or 
decreased in any other ratio such as 1 J to 1 or 3 to 1 by the 
same method. In the case of extremely high ratios, however, a 
change in space factor may be expected and Figs. 19 and 20 
should be consulted. If the current density is increased appreci- 
ably the possibility of overheating should be carefully investigated, 
and it may prove necessary to reduce the power output of the motor. 

In the case of motors over J hp. wound for voltages over 
220 volts or of motors over J hp. wound for more than 110 volts, 
the reactance voltage should be investigated as explained in the 
illustrative design of the i-hp. compound-wound motor. If the 
reactance voltage is too high, one of the following remedies must 
be employed: increase number of commutator bars; reduce output 
of motor; or employ more flux and fewer armature inductors. 

In the case of very low voltage motors in which the brush 
drop is over 5 per cent of the total voltage, the proportions 
should be based on the line voltage less the brush drop when 
computing the armature inductors. 

CASE IV— CHANGE OF SPEED 
Necessary Data. The motor to be rewound is as follows: 

Given: Direct-current motor, | hp., 110 volts, 2000 r.p.m. Shunt wound. 

In good condition. 
Required: Rewind so as to increase speed to 3000 r.p.m. at 110 volts and obtain 

as much power as possible. 



•7 



98 DESIGN OF SMALL MOTORS 

It is of course possible to increase the speed of a motor 
merely by inserting resistance in the shunt-field circuit, but this 
method has the disadvantage that it will not then be possible to 
load the motor to its full capability. This is true because a 
weakened field means a smaller ratio of field ampere turns to 
armature ampere turns and hence poorer commutation, a lower 
breakdown torque, and increased heating. 

To obtain the maximum power output of which a motor is 
capable, it is necessary to rewind the armature if the speed is 
increased. The field may, however, remain the same, since the 
voltage is the same and it is desirable to keep the flux the same. 

Armature. It is evident from the voltage equation 

60 c 

that an increase in r.p.m. should be accompanied by a correspond- 
ing decrease in N since the other quantities are to remain con- 
stant. Thus, in this particular case, an increase in speed from 
2000 to 3000 r.p.m. will mean reducing the number of armature 

2000 2 
inductors to wf^y or — , of the old number. The inductor section 

will be increased to f times the former section. The armature 
current will have to be increased to f times its former value 
before the heating due to PR loss will be the same. Owing to 
the fact, however, that the speed is to be increased, the friction 
losses will be proportionately increased. Moreover, owing to the 
increased frequency in the armature, the iron losses will be in- 
creased. Therefore, in order to keep the original temperature 
rise of the motor, it will be necessary to use a somewhat lower 
current density in the armature winding, in this case probably f 
of the old density. The new armature current will therefore be 
fXf, or t of the old current. The power of the motor will, 
therefore, have been increased to JXf, or 0.281 hp. 

In general, therefore, the number of armature inductors is 
inversely proportional to the speed, and by following this 
method the speed may be increased or decreased in any desired 
ratio. Of course, such changes are to be made only within 
decided limitations. The following are the most important limita- 



98 



I 

tio: 



DESIGN OF SMALL MOTORS 99 



ions and should be checked, especially if the ratio of the speeds 
differs greatly from 1: 

(1) Heating should be about the same. 

(2) Commutation conditions should not be seriously compromised. See 
that the ratio of field to armature ampere turns is not greatly reduced. If the 
voltage is over 110 and the motor is over | hp., check the reactance voltage. 

(3) If the ratio of speeds is great, the armature IR drop and the brush drop 
should be subtracted before taking proportions. In general, the IR drop will be 
a smaller percentage of the total voltage, the higher the speed. 

In the case of a series-wound motor the field ampere 
turns must be kept approximately the same, and the num- 
ber of turns must therefore be changed in inverse proportion to 
the current. In the case of large changes in speed this change in 
the series field will affect the armature voltage, and the problem 
resolves itself into a change in both speed and voltage, even 
though the line voltage is to be the same. The simplest pro- 
cedure in this case is to assume a value of current and find the 
solution by trial. 

CASE V— CHANGE IN DIRECTION OF ROTATION 

Procedure. To change the direction of rotation of a d.c. 
motor, reverse the armature circuit with respect to the field circuit. 
In most cases it is easier to reverse the field terminals, but, if the 
machine is compound wound, care should be taken that the two 
field windings aid each other. If the brushes have been shifted 
to improve commutation, it will be necessary to shift them back- 
ward against the new direction of rotation until a favorable 
operating point is found, as evidenced by improved commutation. 

INDUCTION MOTORS 

Constants of Design. In Case VI it will be assumed that the 
designer has an induction-motor frame in good condition, except 
that the primary winding is burned out and the name plate lost. 
It will be shown how the motor may be rewound and given a 
suitable horsepower rating. The other cases will cover changes in 
voltage, frequency, speed, etc. 

In general, the fundamental constants of design which must 
be held within suitable limits are: mean crest flux density in the 
air gap; ampere conductors per inch of periphery; and peripheral 



99 



100 DESIGN OF SMALL MOTORS 

velocity. Suitable values for these constants are given under the 
particular cases. 

CASE VI 

Necessary Data. The data regarding the motor under con- 
sideration follows: 

Given: Induction motor with squirrel-cage rotor and centrifugal cutout. 

In good condition, except primary windings burned out and name 

plate lost. 
Required: Rewind motor for 110 volts, 60 cycles, single phase, 1800 r.p.m., and 

determine suitable horsepower rating. 

The rotor should be removed and the following dimensions 
carefully measured: 

Stator (Primary) 

Outside diameter of punching 6.6 inches 

Inside diameter of punching 3.8 inches 

Frame length 2.36 inches 

Number of slots 36.00 

Slot width (uniform) 0.25 inch 

Slot depth 0.85 inch 

Useful slot depth . . , 0.80 inch 

Width of slot opening 0.11 inch 

Slot pitch 0.332 inch 

Width of tooth at neck 0.093 inch 

Width of tooth at base 0.23 inch 

Width of tooth at top 0.222 inch 

Depth of core 0.55 inch 

Air gap 0.013 inch 

Rotor (Secondary) 

Outside diameter 3.774 inches 

Length 2.42 inches 

Number of slots (filled with circular bars). . . 25.00 

Slot pitch 0.475 inch 

Diameter of slot 0.327 inch 

Depth slot center below surface 0.2135 inch 

Width of slot opening 0.05 inch 

Depth of core 0.50 inch 

Section of end ring 1 X0.175 0.175 square inch 

As the speed is to be 1800 r.p.m., or 30 revolutions per 
second, there will be two cycles per revolution with 60 cycles 
supply, and hence two pairs of poles, or four poles, will be required. 

Distribution of Flux. Assume a mean gap crest flux density* 
of 25,000 Hues per square inch, and a sinusoidal distribution of 
the flux. In the region of maximum flux density the flux through 

* This density varies from 20,000 to 35,000 for induction motors with nearly closed slots 
under 1 hp., the lower densities being used with the smaller motors and motors having a larger 
number of poles. 



100 



DESIGN OF SMALL MOTORS 



101 



one tooth will be 25000X^X2.36, or 19610 lines. At the instant 

when this occurs in teeth 5 and 6, Fig. 26, the flux in the teeth 
will be as follows: 

Teeth Numbers Per Cent of Maximum Flux in Teeth 

5 and 6 100 19610 

4 and 7 85 16670 

Sand 8 62 12140 

2 and 9 35 6865 

1 and 10 

Total 55285X2, or 110570 

Main Winding. The voltages induced in one quarter of the 

main winding will be as follows: 



Turns 



Teeth Included 



Flux Included 

(From previous 

tabulation) 



Two outer 

Next two 
Next one 
Inner turn 



110570 
98840 
72560 
39220 



Flux XTurns 

221140 

193680 

72560 

39220 



Induced Voltage 
V27r/^iV10-8 

0.590 volt 
0.517 volt 
0.1936 volt 
0.1048 volt 
1.4054 volts 



2,3,4,5,6,7,8,9 

3,4,5,6,7,8 

4,5,6,7 

5,6 

Total 526600 

The above voltage is to be multiplied by four for the four quarters 
and by a factor Ki by which the number of conductors shown in 
Fig. 26 is to be multiplied. Estimate the back-e.m.f. as 96 volts, 
leaving the vector difference between 110 volts and the back- 
e.m.f. as the amount of impedance drop in the winding. Therefore 



or 



1.4054X4Xi^i = 96 
i^i = 17.1 



Make Ki equal 17, giving a back-e.m.f. of 95.6 volts. There will 
thus be 34 conductors per slot in the main winding in those slots 
in which two conductors are shown. The total number of con- 
ductors in the main winding will be 48X17, or 816. The slot 
area will be 0.25X0.8, or 0.2 square inches. 

Allowing for a 0.015-inch slot lining and a slot wedge at the 
top 0.040 inch thick, the remaining available winding space will be 
0.1622 square inch. Assume that one-half of this, or 0.0811 
square inch, will be copper section. The size of conductor will 

thus be -^— — , or 0.00239 square inch. This section comes 

between No. 15 and No. 16, but as a larger conductor than 
0.00239 square inch cannot be employed. No. 16, which is 0.002028 



101 



102 



DESIGN OF SMALL MOTORS 



square inch, will be used. The length of mean conductor will be 
that of a conductor in the slot between teeth 2 and 3. Making 
the allowance for end connections and corners as in the previous 
illustrative designs, the length of mean conductor will be -6.8 
inches, or 0.567 foot. The total length of wire will thus be 
0.567X816, or 463 feet. The resistance of No. 16 is 0.004489 
ohm per foot, making the resistance of the main winding 0.004489 
X463, or 2.08 ohms. 

Magnetizing Current. The primary slot contraction factor 

Wt^-2^ 



will be 



V 



tooth pitch. Substituting gives 



where Wt is tooth width, A is air gap, and p is 
0.222+2 (0.013) 



0.332 
0.748. The secondary slot contraction factor will be 
or 0.948. 



which equals 
0.425+0.026 



0.475 



will therefore be 



The actual flux density in the air gap at crest value 
25000 



-, or 35300 lines per square inch. 



0.748X0.948' 

Fig. 22 is used to obtain the ampere turns per inch and the 
tabulation is as follows: 



Part 

Primary core 
One tooth 
Air gap 

Secondary teeth 
Secondary core 



Flux 

1105704-2 
19610 
19610* 
25500 

100000^2 



Section 

0.55X2.36 
0.093X2.36 

0.094X2.42 
0.5X2.42 



Flux 
Density 

42600 
89400 
35300 
112000 
41300 



Ampere 

Turns 

per Inch 

4 

32 

10960 

200 

4 



Length 

2.0 

0.25 

0.013 

0.125 

1.0 



Ampere 

Turns 

per pole 

8.0 

8.0 

142.6 

25.0 

4.0 



Total 187.6 



The 187.6 ampere turns per pole must be multiplied by a 
factor which will account for the part of excitation supplied by 
the secondary winding. According to the statement in the treat- 
ment of the |-hp. induction motor, this factor for the size of 
motor under consideration will be about 1.85. The ampere turns 
per pole will thus be 1.85X187.6, or 347 at the instant of maxi- 
mum flux and hence maximum magnetizing current. The effective 
value of the magnetizing ampere turns per pole will be 347-^V2, 
or 245.5. The number of turns per pole on the main winding is 

* Under one tooth at crest value. 



I 



102 



DESIGN OF SMALL MOTORS 103 

245 5 
to be 102, and hence the magnetizing current will be ^' , or 2.41 

amperes. As explained in the previous illustrative designs, the 

magnetizing current should be about 0.75 of the total reactive 

2 41 
current. Thus the reactive current will be -— -, or 3.215 amperes. 

A motor of this size should have about 175 in-phase ampere con- 
ductors per inch of periphery. Thus, since there are to be 816 

conductors, the in-phase current will be ' — , or 2.56 

olu 

amperes. The total current will be V3.2152+2.562, or 4.11 

2.56 
amperes. The power factor will be ^—, or 0.623, which is a 

reasonable value. The current density in the primary winding 

4 11 
will be ^ ^ ' ^^^ > or 2027 amperes per square inch, which is 
0.002028 i- F M 

satisfactory. In general, it should be about 2500 in small motors. 

The preliminary assumptions are therefore satisfactory. 

Starting Winding. The starting winding will be arranged as 

in Fig. 27, except that K2 will be chosen so as to make the total 

number of conductors between 50 per cent and 60 per cent of the 

conductors in the main winding. Taking 0.55, the number of con- 

449 
ductors will be 0.55X816, or 449. K2 will therefore be — — , or 

11.2. If K2 is taken as 11, the total number of conductors will 
thus be 11X40, or 440. The length of mean conductor will be 
about the same in the starting winding as in the main winding, 
because the length of the end connections will be greater in the 
starting winding since it is to be wound over the main winding, 
and this will about compensate for the fact that it is nearer the 
slot openings. Using 0.567 foot per conductor, the length of wire 
will be 0.567X440, or 249.5 feet. The resistance of the winding 
should be such as to limit the initial current to about four times 
the normal full-load current. Since the voltage is to be 110, the 

impedance will thus be , or 6.69 ohms. The larger part of 

the impedance will be due to resistance, and the starting winding 
resistance will be taken as 6 ohms. The resistance per foot of 



103 



104 DESIGN OF SMALL MOTORS 

wire must therefore be — -— , or 0.02404 ohm. No. 24 is 0.02567 
249.5 

ohm per foot (cold). The total copper section for the 22 con- 
ductors per slot will be 22x0.0003173, or 0.00698 square inch. 

This will give ample slot room in those slots which contain 
the starting winding only. As the section is to be less than one- 
half that of the main winding, there will be ample room in those 
slots which contain part of each winding, such as the slot between 
teeth 3 and 4- 

Secondary Winding. The secondary winding is assumed to be 
in good condition and it is merely necessary to roughly estimate 
the copper section in order to be sure that the power loss in it 
will be reasonable and thus give the motor a reasonable slip, etc. 

The cross-section per bar is — XO.3272, or 0.084 square inch. 

The total cross-section for 25 bars will be 2.1 square inches. In 
the primary winding the copper cross-section per slot is to be 
0.0811 and there will be 36 slots, and therefore there will be 
0.0811X36, or 2.92 square inches. But as the primary circuit is 
to have somewhat more ampere conductors, this ratio of sections 
is reasonable. In general, for small motors the secondary copper 
section should be between two-thirds and one times the primary 
copper section. The cross-section of the end ring will be 0.175 

square inch. This is equal to the section of ' , or 2.08 bars. 

In general, this section should be equal to the section of about 
one-third the number of bars per pole, and it is thus satisfactory 
in this case. Incidentally it shows that the original winding was 
probably also a four-pole winding. 

Rating. The new windings for the motor have now been 
computed, and it merely remains to decide how much load the 
motor should carry. This involves determination of the losses and 
the resulting temperature rise, which may be estimated from 
Fig. 28. This procedure is given in detail in the previous illustra- 
tive designs and may be worked out if desired. Assume that the 
estimated value of 4.11 amperes total current gives losses resulting 
in a reasonable temperature rise and may therefore be taken as 
the permissible full-load current. 



104 



d 



t DESIGN OF SMALL MOTORS 105 

The efficiency of the motor would probably prove to be 
ut 60 per cent, and the power factor is estimated as 0.623. 
With a voltage of 110 the input to the motor will thus be 4.11 X 
110X0.623, or 281.5 watts. The output will be 0.6X281.5, or 169 

169 
watts, which equals — — , or 0.226 hp., or a little less than J hp. 

The windings for this J-hp. motor have been determined as: 

Main: 816 conductors of No. 16 double cotton-covered wire. Ki = 17; Fig. 26. 
Starting: 440 conductors of No. 24 double cotton-covered wire. ^8^2 = 11; Fig. 27. 

CASE VII— CHANGE OF FREQUENCY 

Effect of Change of Frequency. The usual range of frequency 
is from 25 to 60 cycles. Discussion is understood to be confined 
to this range unless otherwise stated. In changing from one fre- 
quency to another the most essential requirement is to keep to a 
flux density within suitable limits. A low density will reduce the 
maximum torque to a low value, while a high density will result 
in a low power factor and perhaps overheating. Thus a 110- volt 
25-cycle motor, if operated on a 60-cycle source, would have 

about the same flux density in the air gap if the applied voltage 

An 
were increased to 7— XllO, or 264 volts. This is due to the fact 

25 

that the back-e.m.f., which is roughly proportional to the applied 

voltage, is directly proportional to the frequency and the flux 

density. In fact, a 110-volt 25-cycle motor wiU usually operate 

satisfactorily, but with somewhat less torque output, on a 220-volt 

An 
60-cycle circuit; will run at — of the former speed; and will 

develop increased power owing to the increased speed. 

Therefore, in general, an induction motor rated at a certain 
frequency may be operated at another frequency, provided the 
applied voltage is changed in approximately direct proportion to 
the changed frequency. The torque will remain approximately 
constant, but the power output and the speed will be directly 
changed with the changed frequency. The above statements apply 
of course only within certain limits, and in any case the heating, 
power factor, and safe load may be altered. Nevertheless it is often 
possible to make use of second-hand motors by applying this rule. 



106 



106 DESIGN OF SMALL MOTORS 

Necessary Data. If the voltage cannot be changed or if it 
would not be satisfactory to use the resulting speed with the 
changed frequency, the motor must be rewound. An example of 
a 60-cycle motor rewound to run on 25 cycles follows: 

Given: |-hp., 60-cycle, 110-volt, 1800-r.p.m., single-phase induction motor in 

good condition. 
Required: Rewind to operate on 25 cycles, 110 volts, 1500 r.p.m. synchronous 

speed. Determine suitable power rating at new frequency. 

For convenience, it will be assumed that the given motor is 
the J-hp. induction motor, frame No. 165, which was worked out 
in detail in the previous illustrative design, and reference is under- 
stood to be made to this design for dimensions and other data. 
This design employed a four-pole winding, but in the case of the 
rewound motor a synchronous speed of 1500 r.p.m., or 25 revolu- 
tions per second, must be obtained, which requires one pair of 
poles for 25 cycles. The new winding must therefore be a two- 
pole winding. It is evident that if the gap density remained the 
same, the flux in the primary and the secondary cores would be 
doubled. This would saturate the iron and give a very bad 
power factor and also large iron losses. It will therefore be neces- 
sary to reduce the gap density somewhat. 

Determination of Gap Density. In view of the fact that a 
frame intended for a four-pole design is being used, the iron sec- 
tion in the core will be ^mailer than it should be and will therefore 
determine the flux which may be employed. Ordinarily a mean 
gap density of about 30,000 lines could be used with a two-pole 
design. In this case, however, on account of the smaller iron 
cross-section, a good many ampere turns will be lost in the iron 
parts. Therefore a mean gap density of 25,000 lines will be a 
good value to try. Assuming a sinusoidal distribution of flux, the 

crest value of mean gap density will be — times the average value. 

The area of one pole (one semicircumference) will be J X 6.3 12 
inches diameter XttX 2.5 inches length, or 24.8 square inches. The 

2 
flux per pole will thus be -X 25000X24.8, or 394500 lines. 

TT 

Main Winding. The primary windings will be distributed in 
a way similar to that shown in Figs. 30 and 31 for the 60-cycle 



106 



I 



DESIGN OF SMALL MOTORS 107 



design, except that they will be two pole. The following tabula- 
tion summarizes the winding data for both main and starting 
windings : 





Conductors 


Conductors 


Slot Number 


Main Winding 


Starting Winding 


1 





2K^ 


2 





2K2 


3 


K, 


2K2 


4 


Kr 


2K2 


5 


2K, 


2K2 


6 


2K, 


2K2 


7 


2Ki 


2K2 


8 


2Kx 


K2 


9 


2Ki 


K2 


10 


2Ki 





11 


2K, 





12 


2K, 





13 


2K, 





14 


2K, 





15 


2K, 





16 


2Ki 


K2 


17 


2K, 


K2 


18 


2K, 


2K2 


19 


2Kx 


2K2 


20 


2K, 


2K2 


21 


K, 


2K, 


22 


K, 


2K2 


23 





2K2 


24 



Total per pole 36 Ki 


2K2 




32 ^2 



Ki and K2 are the constants by which the number of conductors 
listed for each slot is to be multiplied. 

If a diagram such as Fig. 30 were made and the number of 
flux turns per pole computed by the procedure outlined in the 
60-cycle design, it would be found that there are to be 5,275,000 
Ki flux turns per pole for the main winding. The e.m.f. induced 
in these flux turns will be \/27r/<|)iV10-8, or \/2X7rX 25X5275000 
XiXlO"^ which equals 5.875 Ki volts. The total back-e.m.f. for 
two poles will thus be 2X5.875 Ki, or 11.75 i^i. If a back-e.m.f. 
of 100 volts is assumed 

^^-1L75-^-^ 

If Ki is taken as 8, the mean gap crest density will be increased. 

8 5 
to -^X 25000, or 26550 lines per square inch, and the total 

8 

flux to ^X 394500, or 419000 lines. The maximum number of 

8 



107 



108 



DESIGN OF SMALL MOTORS 



conductors per slot will thus be 2Ki, or 16 conductors, and the 
total number of conductors in the main winding will be 576. 
The number of turns per pole will be 18i^i, or 144 turns. 

The copper section per slot available for the main winding 
was found to be 0.0477 square inch in the 60-cycle design and will 

0.0477 



be the same in this case. The conductor section will be 



16 



or 0.00298 square inch. No. 15 is 0.002558 square inch, and No. 
14 is too large; therefore No. 15 will be used. The length of 
mean conductor will be approximately that of a conductor in slot 
No. 8 and will be the sum of the frame length, the arc, and the 
end allowance, which equals 2.5+4.59+1, or 8.09 inches, or 0.674 
foot. The total length of wire in the main winding will be 0.674 
X576, or 388 feet. The hot resistance of No. 15 is 0.00356 ohm 
per foot, making the resistance of the winding 0.00356X388, or 
1.38 ohms. 

Magnetizing Current. Before proceeding further, the mag- 
netizing ampere turns should be determined to see whether it will 
be necessary to revise the estimate of gap density. 



Part 



Flux 



Section 



Ampere Ampere 

Flux Turns Turns 

Density per Inch Length* per Pole 



Primary core 


419000- 


-2 0.92X2.5 


91000 


30 


1.5 


45.0 


One tooth (maximum 














density) 


27450 


0.15X2.5 


73200 


13 


0.75 


97.5 


Air gap (one tooth) 


27450 


0.292X2.5 


37600 


11780 


0.012 


141.3 


Secondary core 


391000- 


-2 0.8X2.5 


97600 


60 
Total.'. 


3.0 


180.0 
. 463.8 



The value 37,600 for flux density in the air gap is increased above 
the mean gap density of 26,550 by slot contraction. The slot 
contraction factors for this frame were computed in the 60-cycle 
design. 

The value 464 for the ampere turns per pole must be multi- 
plied by 1.9 to allow for the secondary excitation. The total 
excitation will be 1.9X464, or 882 ampere turns maximum. The 



The number of 



882 
effective value will be —^, or 623 ampere turns. 

turns per pole on the main winding is to be 144. The effective 

* Length over which stated density exists. 



108 



DESIGN OF SMALL MOTORS 109 

623 
value of the magnetizing current will thus be — — , or 4.325 

amperes. The ratio of the magnetizing current to the total 

reactive component will be large in this case, probably 0.85, mak- 

4 325 
ing the total reactive current ' , or 5.09 amperes. Choosing a 

0.85 

value of 175 for the in-phase ampere conductors per inch of periph- 
ery, the in-phase current will be approximately — — — ' 

576 

or 4.3 amperes. The total current will be v4.32+5.092, or 6.66 

4 3 
amperes, and the power factor will be — ^, or 0.645, which is 

6.66 

satisfactory. The current density in the main winding will be 

-— ^-— — , or 2600, which is satisfactory. 
0.002558 

Starting Winding. The starting winding will have about 55 

per cent of the number of conductors of the main winding. 

Referring to the previous winding tabulation 

0.55X576 = 64^2 

and K2 equals 4.95. If K2 is taken as equaling 5, there will be 
5X64, or 320 conductors, in the starting winding. The resistance 
of this winding will be sufficient to limit the initial current to 
about four times normal full-load current. Its impedance will 

thus be - — -— -, or 3.49 ohms. As most of the impedance will be 

resistance, a resistance of 3.25 ohms will be employed. The length 

of mean conductor for the starting winding will be approximately 

0.75 foot, making the total length of wire 0.75 X 320, or 240 feet. The 

3 25 
resistance per foot will thus be -^^, or 0.01355 ohm. No. 21 is 

0.0128 ohm per foot (cold) and is the nearest size. Therefore 
No. 21 double cotton-covered wire will be used for the starting 
winding. 

In a slot such as No. 5, which contains conductors of both 
windings, the copper section will be: 16 conductors of No. 15, 
16X0.002558, or 0.0409 square inch; and 10 conductors of No. 21, 
10X0.0006363, or 0.006363 square inch; making a total of 0.047263 



109 



no DESIGN OF SMALL MOTORS 

square inch. The available winding space per slot was determined 
in the 60-cycle design as 0.1122 square inch, which is more than 
double the copper section. It is therefore certain that there will 
be ample room for the windings. 

Secondary Winding. The same squirrel-cage secondary will 
be used, but the current in the end rings will be double the former 
value. This will mean high copper losses in the secondary and a large 
slip. On the other hand, the starting torque will be improved. 
If the particular requirements of this case did not call for a high 
starting torque but did require a high breakdown torque, the old 
end rings would be replaced with rings of double the thickness. 

Rating. The losses may be computed in the same way as in 
the 60-cycle design, except that the iron losses for 25 cycles are 
computed by the formula 

watts per cu. in. = 1.3 ( -— ^ 
\10^ 

The total losses for this 25-cycle design will probably prove to be 
greater than in the case of the 60-cycle design. The temperature 
rise may then be computed in the same way and a decision made 
as to whether the estimate of 7.89 amperes w^ill be satisfactory. 
In case this value proves permissible (as is probable), this motor 
will develop J hp. or a trifle more, since the in-phase current 
would be about 6 amperes as compared with the previous value of 
5 amperes for the 60-cycle design. 

The windings of the J-hp. motor will be as follows: 

Main winding: 576 conductors of No. 15 double cotton-covered wire dis- 

tributed as per tabulation. 

Starting winding: 320 conductors of No. 21 double cotton-covered wire dis- 
tributed as per tabulation. 

Secondary winding : Same squirrel cage as in previous 60-cycle design, except 
end ring is to be of double thickness. 

CASE VIII— CHANGE OF VOLTAGE 
Voltage Ratio. To change the voltage of an induction motor, 
keeping all other characteristics the same, it is merely necessary to 
change the number of turns in the primary winding in direct pro- 
portion to the change in voltage. Thus, if the voltage is doubled, 
the number of turns must be doubled, or, if the voltage is reduced 
to one-half value, the number of turns must be reduced to one- 
half the former value. 



10 



DESIGN OF SMALL MOTORS 111 

Owing to this simple relation, the new windings may be 
determined by counting and tabulating the number of conductors 
of both main and starting windings in each slot and noting the 
scheme of connections. The number of conductors in each slot is 
then multiplied by the ratio of the new to the old voltage and the 
motor rewound, using the old scheme of connections. In case the 
number of new conductors per slot is fractional the nearest whole 
number may be taken, but an effort should be made to have the 
total number of new conductors in both main and starting wind- 
ings close to the computed value. The cross-section of the new 
conductors compared with the old should be in inverse proportion 
to the changed voltage, so that the ratio of the total copper sec- 
tion to the slot section remains the same. In case the desired 
section comes between two sizes of wire, it is preferable to use 
the larger size if there is sufficient slot room, but in any case the 
number of turns should be kept as close as possible to the value 
computed from the voltage ratio. 

Change without Rewinding. It is sometimes possible to 
change the voltage without rewinding in case the voltage ratio is 
evenly divisible into the number of poles. For example, consider 
a 220-volt four-pole motor wound with all the turns in series (as 
in Fig. 26). By cutting the middle point of the winding it may 
be divided into two parts, which may be reconnected in parallel, 
so that current flows in the same relative direction in each part. 
The motor w^ill then operate as before on 110 volts. Similarly 
the motor may be made to operate on one-fourth of the original 
voltage by dividing it into four parts and reconnecting it in paral- 
lel. A good method of locating the proper point in the winding to 
make the cut is to apply a low d.c. voltage (which will not over- 
heat the motor) across the entire winding and by means of a 
needle point prick through the insulation of the end connections 
and measure the voltage between one terminal and the needle. 
Thus the midpoint of the winding will be that point which shows 
one-half the total applied voltage. 

CASE IX— CHANGE OF SPEED 

Number of Poles. To change the speed of an induction 
motor without changing the frequency requires a change in the 



111 



112 DESIGN OF SMALL MOTORS 

number of poles, and hence rewinding. The number of pairs of 
poles is frequency divided by revolutions per second (synchronous 
speed). It is thus obvious that the number of speeds is limited 
for a given frequency, since it is not practical to use over eight 
poles on small motors. The following tabulation shows the various 
combinations which are practical for small motors for various fre- 



quencies : 












25 Cycles 
Poles R.P.M. 


40 Cycles 
Poles R.P.M. 


60 C^ 
Poles 


CLE8 

R.P.M. 


133 Cycles 
Poles R.P.M. 


2 1500 
4 750 


2 2400 
4 1200 
6 800 


2 
4 
6 

8 


3600 

1800 

1200 

900 


4 
6 

8 


4000 
2666 
2000 



Mean Gap Density. In order to obtain reasonable power 
factors, the mean gap density must be reduced in case the number 
of poles is increased. If the number of poles is reduced, the flux 
in both primary and secondary cores is correspondingly increased 
for a given gap density. Unless there has been an excess of core 
section provided in the original design, it is thus necessary to use 
values of gap density sufHciently low to avoid saturation of the 
iron. In a two-pole motor the mean gap density at crest value 
should be from 25,000 to 35,000 lines per square inch. The 
higher values are used on the larger motors. In an eight-pole 
motor this density should be between 15,000 and 20,000. In 
general, the higher the density used in a given frame the lower 
the power factor, but the greater the permissible load. 

After deciding the number of poles, the design may be 
worked out by the same methods given in detail previously. 

Approximate Method of Computing Winding. . It is possible to 
compute an approximately correct winding for a given case by 
proportion, using the old winding as a basis. As it often happens 
that the frequency and the voltage as well as the speed are 
changed, the following procedure takes this fact into account. 
Denoting the old winding by the subscript 1 and the new" winding 
by the subscript 2, we have: 

(turns per pole) o = (turns per po\e)iX-z^X-FrXj 

7(2)2 xSi J2 

where /3 is gap density, E is line voltage, and / is frequency. 

In this formula the same distribution of windings is assumed. 
Owing to the large number of elements involved, the formula 



112 



DESIGN OF SMALL MOTORS 113 

should be taken merely as a rough guide, and after computing the 
new winding it should be checked by working out the complete 
design according to the method used in the illustrative designs. 

AUTOMOBILE STARTING AND LIGHTING 
EQUIPMENT 

Classification. Automobile starting and lighting systems may 
be classified under one-, two-, and three-unit systems, all of which 
employ a storage battery. The one-unit system is a system in 
which one dynamo performs the three functions of starting motor, 
lighting generator, and ignition generator. In what is usually 
termed the two-unit system, one dynamo acts as a starting motor 
and a second dynamo as a generator which supplies lighting and 
ignition current. Fig. 34 shows a typical two-unit outfit made by 
the Remy Electric Company. In the three-unit system separate 
machines are employed for starting, lighting, and ignition. Another 
system employing one machine for starting and lighting and a 
second (usually a magneto) for ignition only is sometimes called a 
one-unit system and sometimes a two-unit system. As magneto 
design involves radically different principles, this article will deal 
only with the design of the starting motor, the generator for light- 
ing and battery charging, and the combined machine which acts 
as both starting motor and generator. 

STARTING MOTOR DESIGN 
Requirements. In Fig. 35 is shown a typical starting motor 
made by the Delco Company and used on the 9-N Continental 
engines. In a starting motor the determining requirements of the 
design are the breakaway torque and the running torque at crank- 
ing speed. The present practice is to use an engine speed in the 
neighborhood of 100 r.p.m. In the past, speeds as low as 30 r.p.m. 
were employed. Starting motors are series wound, so that it is 
convenient to rate them in terms of torque rather than of horse- 
power. Fig. 36 shows the breakaway and running torque as a 
function of the total cylinder displacement of the engine at a 15 
to 1 gear ratio between the starting motor and the engine. It should 
be noted that the engine is specified in cylinder displacement 
instead of horsepower since its horsepower depends upon its speed 



113 




114 



DESIGN OF SMALL MOTORS 115 

as well as" upon its cylinder displacement, while the breakaway 
torque depends merely on cylinder displacement. At usual engine 
speeds, however, 200 cubic inches displacement would correspond 
to 20-25 hp. 

Low Voltages Employed. The design of the starting motor is 
influenced by the characteristics of the storage battery. The lead 
battery is used almost exclusively rather than the Edison battery, 
because the former is somewhat cheaper and has a lower internal 
resistance; hence it gives a greater current output for developing 
breakaway torque. With variation in outdoor temperature the 
internal resistance increases and the ampere-hour capacity decreases, 




Fig. 35. Typical Starting Motor, Used on 9-H Continental Motor 
Courtesy of Dayton Engineering Laboratories Company , Dayton, Ohio 

and thus the most severe condition for the battery occurs in winter 
service. It is customary to specify batteries of sufficient size to 
develop the required breakaway current for five seconds at a 
temperature of 10° F. at a voltage, at the battery terminals, of 
not less than 1.5 volts per cell. There are in use 3-, 6-, and 12- 
cell systems, which correspond to what are known as 6-, 12-, and 
24-volt systems. The 6-volt system is used most frequently, 
because there are fewer cells to care for and the energy stored per 
pound is slightly greater with the larger sizes of cells. 

On the other hand, the efiiciency of the starting system is 
somewhat grea.ter with higher voltages owing to the smaller per- 
centage of IR drop in the leads and brushes. It has been claimed 



115 



116 



DESIGN OF SMALL MOTORS 



that the 12-volt system is the best compromise, although, up to 
the present writing, 6-volt systems are more generally in use. 
But in all systems the voltage is far below that used in other 
power service, and hence, in general, the design of automobile 
starters and generators is confined to high-current low- voltage 
machines. This requires large brushes having a minimum voltage 



I 



30 

28 

26 

Z4- 

22 

20 

Id 

It 

14 

12 

10 

8 

6 

4 



1 


n 


n 


~^ 


^ 


n 


n 




n 


"I 




















'AUTOMOBILE STARTERS 














4 


/ 


TORQUE REQUIREMENTS 












/ 


/ 


































y 


r 


































/ 


r 


































/ 




































/ 


^ 


































/ 


nO 
































/ 


^0^ 
































/ 


A 


































/. 


^^ 


1^ 
































/M 
































/ 


t 


V 
































/ 


X 
































/ 


^ 




































f 




































} 






































1 


■ 


































/ 






































/ 






( 


"jEAR RATIO BETWEEN 










/ 


f 






^ 


5TAR 


TER AND 


ENGINE 










/ 














15 TO 1 
















_j 


r 




































/ 






























































lie 


























^/^/'^/6 


^TO^^y\"- 


- 




















4 


^L/ 


firl 




i- 


^ 




— 




















i 




^ 


"^ 


'\ 
























X 


^ 










































































100 200 300 400 500 600 TOO 800 900 I000\ 
TOTAL CYLINDER DISPLACEMENT IN CU IN. 

Fig. 36. Torque Requirements for Automobile Starters 



drop. The brush friction losses are thus greater, but the reactance 
voltage is so low that there is relatively little trouble with spark- 
ing or other commutator trouble, provided the machine is given 
reasonable care. In starters the time of operation is so short that 
the heating is not usually sufficient to make it a factor in the 
design of the machine. 



116 



DESIGN OF SMALL MOTORS 117 

STARTER FOR ENGINE OF 300 CUBIC INCHES DISPLACEMENT 
Requirements. The starter to be designed must meet the 
following specifications: 

{Engine 300 cubic inches displacement, cranking speed 100 r.p.m. 
To operate from 3 cells of lead battery with |-volt maximum 
drop in battery leads. 

General Features. Referring to Fig. 36, it is found that 
with 300 cubic inches engine-cylinder displacement the motor 
must produce 16 foot pounds breakaway torque and 2.5 foot 
pounds running torque at 1500 r.p.m., assuming a 15 to 1 gear 
ratio, which is frequently employed. This is equivalent to 

2.5 X27rX 1500 c^^^A x. 17 • • A • 

, or 0.714 hp. J^rom previous experience, a designer 

ooUUU 

would select the size of frame shown in Fig. 37. It will be made 
four pole in order to reduce the cost, weight, and space occupied. 
This is in accordance with the general rule that, if it is desired to 
employ a large flux (in order to get large torque), the yoke section 
must be very large if the design is made two pole. Another dis- 
tinctive feature of the design will be the use of open slots of uni- 
form width. This is to allow the use of form- wound coils. In 
order to hold the coils in these open slots, the completed armature 
will be bound with phosphor bronze wire in two channels ye inch 
wide countersunk in the armature surface yq inch from each end 
of the armature. The pole pieces are to be made up of punchings 
riveted together. They are to be removable and will' be held in 
place by machine screws | inch in diameter countersunk in the 
outside of the frame and extending into threaded holes in the pole 
pieces. The armature will be built up of punchings in the usual 
way. 

Armature Winding. The armature winding is computed as 
follows: Assume an effective mean pole-face flux density* in the 
air gap of 30,000 lines per square inch. As the pole arc will be 2 
inches and the net effective length 3.25 inches, the flux per pole 
will be 30000X2X3.25, or 195000 lines per pole. At full cranking 
speed the voltage of the battery will be about 1.9 volts per cell, 
or 3X1.9, which equals 5.7 volts. Assume a back-e.m.f. of 4 



♦ This density will vary from 20,000 to 35,000 for armatures from 2 to 6 inches in diameter 
with open slots, and from 25,000 to 45,000 for closed slots. 



117 



f!il| 



118 



DESIGN OF SMALL MOTORS 



volts, leaving 1.7 volts for the IR drop in the brushes, series 
field, and battery leads. The number of inductors will then be 

.XWXpathsXGO ^ 4XWX4X6Q_^3^ .^^^^^^^^ 
<I>X poles Xr.p.m. 195000X4X1500 

82 
The number of inductors per slot will be — , or 4.32. The ordi- 
nary form of armature winding requires an even number of induc- 
tors per slot. Therefore 4 inductors per slot will be employed, 

AUTOMOBILE STARUNO 
MOTOR 

QR055 L£N6Tti = 3.875" i 

NET EFFECTIVE LEN6TH 1 

= J. 25'< ■ 




MR 6AP ,025" 
SLOT DEPTH = 70" 
5L0T WIDTH = 22" 
NO OF Tt^TH = 19 
WIDTH OF TOOTH NECK- 1 52" 



5CALE: HALF 5/Z£ 

Fig. 37. Details of Automobile Starting Motor 



4 32 
increasing the mean pole-face density to -^X 30000, or 32400; 

however, 32,000 lines will be used. The total number of inductors 
will be 4X19, or 76. 

The slot area from Fig. 37 is to be 0.7X0.22, or 0.154 square 
inch. Allow a slot lining 0.015 inch thick and a clearance of y^ 
inch between the top of the winding and the periphery of the 
armature. This will leave a winding space 0.6225X0.19, or 0.1182 
square inch. Assume that 55 per cent of this will be copper 
cross-section, giving 0.55X0.116, or 0.06385 square inch. The 

inductor section will be — ^— , or 0.01596 square inch. In order 



118 



DESIGN OF SMALL MOTORS 



119 



to permit the inductors to fit well in the slots, employ such a 
rectangular section that there will be two side by side and two 
deep. If y is the uniform thickness of inductor, insulation, and w 
the width and h the height of the inductor section 

22/+^ = iX0.19 
22/+/i=iX0.6225 
76^/^ = 0.01596 

Solving these equations gives to equals 0.058 inch, h equals 0.275 
inch, and y equals 0.0183 inch. In order to allow a little extra 

AUTOMOBILE STARTING MOTOR ARMATURE yVINDlNS 




+ 
^ T 



Fig. 38. Armature Winding for Automobile Starting Motor 

space for sliding in the inductors, a cotton serving insulation 0.01 
inch instead of 0.0183 inch thick will be employed. The winding 
will be arranged as shown in Fig. 38. The length of mean 
inductor will be the sum of the gross armature length, the end 
connection, and the corner allowance, which equals 3.875+2+1, 
or 6.875 inches. Since there are to be 76 inductors and the 
inductor section will be 0.01596 square inch, the total volume of 
copper in the armature will be 76X6.875X0.01596, or 8.34 cubic 
inches. The total length of wire in the armature will be 76 X 
6.875, or 522 inches. The specific resistance of copper is 0.679 X 
10"® ohms per cubic inch. The resistance of the armature, which 



119 



120 



DESIGN OF SMALL MOTORS 



is to have four paths in parallel of 

130.5 



522 



inches length each, will 



thus be iX0.679Xl0-^X- 



or 0.00139 ohm. 



0.01596' 

Field Winding. The ampere turns required for field excita- 
tion must now be computed. An important detail in this connec- 
tion is the computation of the slot contraction factor in a machine 
having open slots. The slot contraction factor accounts for the 
fringing of the lines of flux around the teeth corners and is the 
ratio of the distance along the tooth face which is effective as a 
flux path to the tooth pitch. Where the air gap is small compared 

Wt-\-2A 



to the slot opening, the slot contraction factor is 



In this 



case the width of the tooth face Wt will be 0.3925 inch; the air- 
gap length A will be 0.023 inch; and p, or the slot pitch, will be 

0.6125 inch. Substituting these values gives — — ^rprrp^ , or 

0.6125 

0.716. Since the mean pole-face density is to be 32,000, the 

actual flux density across the effective portion of the air gap will 

, 32000 ..„^^,. . , 



"^ 0.716' "' 


-Xll \J\J llllCO 


^\:.x ovj[Lia;i\^ X. 


11V.1X. 

Flux 




Ampere 
Turns 


Ampere 
Turns 


Part 


Flux 


Section 


Density 


Length* 


per Incht 


per Pole 


Yoke 


200000 -^ 2 


0.282X6.00 


59000 


1.5 


14.3 


21.43 


Pole core 


220000 


1X3.875 


56800 


0.5 


13.5 


6.75 


Air gap 


195000 




44700 


0.023 


14000.00 


322.00 


Teeth 


195000 


1.865 


104600 


0.25 


97.0 


24.25 


Armature core 


190000-^2 


0.375X3.875 


65400 


0.875 


10.0 


8.75 



Total 383.18 

To allow for armature reaction add 25 per cent, making 1.25 X 
383.18, or 479 ampere turns per pole. Before proceeding further, 
the probable value of the ratio of field ampere turns per pole to 
armature ampere turns per pole should be determined. Since the 
horsepower will be 0.714 and the voltage 6, if the efficiency of the 
motor at running speed is estimated as 65 per cent, the current 
0.714X746 



will then be 



6X0.65 



or 136.7 amperes. The current in each 



* Length over which flux density is considerable, 
t See Fig. 22. 



120 



DESIGN OF SMALL MOTORS 121 

inductor will be 136.7-^4, or 34.2. The number of armature 

turns per pole will be iX|X76 inductors, or 9.5. The armature 

ampere turns per pole will thus be 9.5X34.2, or 325. The ratio of 

field ampere turns per pole to armature ampere turns per pole will 

479 
thus be , or 1.472. This value should be between 1 and 1.5, 

and therefore the design is satisfactory. 

The field winding will, of course, be a series winding, and, 

using the estimated current, 136.7 amperes, and 479 ampere turns, 

479 
the turns per pole will be , or 3.51. It is impossible to 

know exactly how many turns are necessary until the efficiency of 
the motor has been computed. As a preliminary measure, suppose 
that two diametrically opposite poles are wound with four turns 
per pole and the other two with three. This will make it possible 
to proceed without revising the other preliminary assumptions. 
The winding space available for the field coils will be, according 
to Fig. 37, 0.4X1 inch coil section. Allowing a wrapping 0.03 
inch thick, the winding space will be 0.34X0.94, or 0.32 square 
inch. Assume that 0.6 of this will be copper section, or 0.6X0.32, 
which equals 0.192 square inch. The conductor section will thus 

be ^— — , or 0.048 square inch. A ribbon 0.3 inch wide and 0.16 

inch thick will be used. If the corner allowance is made 2 inches, 
the length of mean turn will be 2 (3.875+1+2), or 13.75 inches. 
The volume of copper will be 14X13.75X0.048, or 9.24 cubic 
inches. As there will be 14 turns, the total length of wire will be 
14X13.75, or 192 inches. The resistance will be 0.679 X 10"^ X 

1Q9 

^, or 0.002715 ohm. 

Breakaway Torque. It is now possible to determine the 
breakaway current which the motor will draw. The battery volt- 
age will drop from 6 to about 3.5 volts if the battery is nearly 
discharged. It is customary to allow a drop in the battery leads 
of J volt, leaving 3 volts at the motor terminals. The voltage 
drop in the brushes at breakaway current may be estimated as 
0.3 volt, leaving 2.7 volts across the armature and the field. 
With an armature resistance of 0.00139 ohm and a field resistance 



121 



122 DESIGN OF SMALL MOTORS 

2.7 



of 0.002715 ohm, the breakaway current will be QQQ^3g , q 992715' 

or 658 amperes. The corresponding inductor current will be 658 
-^4, or 164.5 amperes. From the fundamental relations of current 
and flux, a side push of 1 dyne is obtained from a wire 1 centi- 
meter long carrying a current of 1 abampere in a field of unit flux 
density. Reducing this formula to practical units, the following 
equation results 

P = 0.SS5X MINI X10-' 

where P is pounds, /3 lines per square inch, / inches, N the num- 
ber of inductors, and / the inductor current. Substituting the 
numerical values in this case 

P = 0.885X42500X3.25X(fX76)Xl64.5Xl0-^ = 101.81b. 

The value of M was increased 33 per cent from 32000 in order to 
allow for the greater flux density at the breakaway current. 
Although the field ampere turns are much greater than at full 
speed, the flux does not increase proportionally, owing to satura- 
tion. In practice an experimental saturation curve is available to 
a designer, but an approximate saturation curve may be calculated 
by the method here employed, using various values of flux. The 
number of inductors in the above formula w^as taken at two- 
thirds of 76, since only two-thirds of the total number are in the 
field at any instant. 

The 101.8 pounds will be exerted at the surface of the arma- 
ture, which has a radius of 1.85 inches. The breakaway torque 

1 85 
will thus be 101.8 X^-^, or 15.7 pound feet, when the battery is 

in the worst condition. This is sufl&ciently close to the value of 
16 pound feet called for by Fig. 36. ^| 

Losses. The original assumptions have now been checked, 
and consideration may again be given to the full cranking speed 
condition. First determine the efficiency with the design as itfll 
now stands. In computing the iron losses in the teeth, it is nec- 
essary to take account of the fact that the flux density varies 
from point to point. First divide the tooth into two parts by an 
arbitrary line at one-half slot depth, and then assume that the 
flux density in the outer half will be that at the tooth surface and 



122 



DESIGN OF SMALL MOTORS 123^ 

the density in the inner half that at the tooth neck. The volume 

0.35 
of iron in the upper halves of the nineteen teeth will be 19 X-^^ 

X (0.3925+0.2725) X3.875, or 8.55 cubic inches, and the weight 
will be 2.38 pounds. From the saturation-curve table the flux 
density will be 44,700, and, from Fig. 21, 2.9 watts per pound is 

obtained at -7777- X 2, or 50 cycles. The iron loss in the outer 
60 

halves of the teeth will thus be 2.9X2.38, or 6.91 watts. The 

0.35 
volume of iron in the inner halves of the teeth will be 19 X—^ 

X (0.2725+0.152) X3.875, or 5.465 cubic inches, which will weigh 
1.52 pounds. At 104,600 lines per square inch, 12 watts per 
pound will be obtained. The iron loss in the inner halves of the 
teeth will thus be 12X1.52, or 18.24 watts. The volume of 
active iron in the armature core will be 0.375 Xl.8757rX 3.875, or 
8.55 cubic inches, which will weigh 2.375 pounds. At 65,400 lines 
per square inch, the loss will be 5 watts per pound, or 5X2.375, 
or 11.87 watts. The total iron losses will therefore be 6.91 + 18.24 
+11.87, or 37.02 watts. 

The brushes will be metallized carbon such as the Le Car- 
bone Company's Metal No. 2 brush, which has a voltage drop 
of only 0.25 volt per pair of brushes with a coefficient of friction 
of about 0.2. It will be well to choose a brush current density 
at running speed of 150 amperes per square inch. The current 

per brush will be — -^, or 68.35 amperes, giving a brush area of 

' , or 0.456 square inch. Therefore a brush IX^ will be used. 
loU 

The commutator will be If inches in diameter and \\ inches 

long and will have nineteen bars. The peripheral speed at 1500 

r.p.m. will be X7rXl500, or 639 feet per minute. The 

brush pressure will be 2.5 pounds per square inch, so that the 

power loss due to brush friction in all four brushes will be 

4X0.5X2.5X0.2X639X746 ,, .^ ^^ rpi 1 1 
, or 14.43 watts. The brush contact 

ooUUU 

loss will be 136.7 amperes XO.25 volt, or 34.2 watts. 



123 



124 DESIGN OF SMALL MOTORS 

The bearing loss with ball bearings is difficult to compute. 
If the bearings are operated with light oil the loss will be very 
small; if, however, they are filled with grease, the churning of the 
grease will cause considerable loss. Allow the same loss as for a 
straight bearing, which would be about 25 watts. On the basis 
of the estimated current of 136.7 amperes the field PR loss will 
be 136.72X0.002715, or 50.4 watts. The armature PR loss will 
be 136.7^X0.00139, or 26 watts. 

The complete table of losses will be as follows: 

Field PR loss 50.40 watts 

Armature PR loss 26.00 watts 

Brush contact loss 34.20 watts 

Iron loss 37.02 watts 

Brush friction 14.43 watts 

Bearings 25.00 watts 

Total 187.05 watts 

Summary. The power output of the motor will be 0.714 
horsepower, or 0.714X746, which equals 532.7 watts. The effi- 

532 7 
ciency will be 533^^^^37.05 ' °' ^^ P'' """*• 

The shop information on this starter will be as follows: 

Mechanical Data 

Frame and punchings as in Fig. 37. 
Electrical Data 
Field Winding: Alternate poles wound with 3 and 4 turns respectively of 
double cotton-covered copper ribbon 0.3X0.16 section. 
Connect in series with armature. 
Armature Winding: 4 inductors per slot of copper ribbon 0.275X0.058 section 
insulated with 0.01-inch cotton serving. 

Example. Work out and tabulate information on a 12-volt starter meeting 
the same requirements as the motor just designed. 

STARTER FOR ENGINE OF 200 CUBIC INCHES DISPLACEMENT 
Design Data. The appearance of one of several types of 
automobile starters made by the Westinghouse Electric & Manu- 
facturing Company is shown in Fig. 39. The tabulation following 
is a description of one frame size. No. 300, which has the per- 
formance characteristics shown in Fig. 40. This motor is intended 
for coupling through a 25 to 1 gear. It produces 6 pound feet 
breakaway torque and 1.1 pound feet running torque at 2500 
r.p.m. Allowing for the higher gear ratio, Fig. 36 shows that 
this motor would be suitable for use with an engine of 200 cubic 
inches total cylinder displacement. 



124 




Fig. 39. Starting Motor Taken Apart 
Courtesy of Westinghouse Electric <& Manufacturing Company 



6000 



5000 



4000- 



i 



3000 



2000 



WOO 




50 100 150 200 

AMPERES INPUT TO MOTOR 

Fig. 40. Approximate Performance of Automobile Starting Motor 
Courtesy of Westinghouse Electric & Manufacturing Company 



125 



126 DESIGN OF SMALL MOTORS 

The following tabulation gives the principal design data: 

Armature 

Diameter 3.25 inches 

Gross length 1.625 inches 

Net length 1.50 inches 

Number of slots 29.0 

Slot pitch 0.3514 inch 

Slot width (imiform) 0.118 inch 

Slot depth 0.491 inch 

Slot opening 0.118 inch 

Thickness of punchings 0.0281 inch 

Diameter of shaft 0.875 inch 

Number of commutator bars 29.0 

Diameter of commutator 1.75 inches 

Length of commutator 1.4375 inches 

*Number of inductors per slot 2.0 

Inductor section 0.0808 inch XO. 175 inch 

Field 

Number of poles 4.0 

Pole arc 1.5 inches 

Field coils (on two poles only) 7 turns per coil 

Field conductor i^ inch X h inch 

CHARGING GENERATOR DESIGN 
Classification. The design of automobile charging generators 
differs from that of ordinary generators, as they must operate on 
variable speed. The principal types may be classified according 
to the following methods by which the current output is regu- 
lated: (1) compound differential field; (2) third-brush method; 
(3) mercury-tube regulator; (4) voltage vibrator controlling shunt 
field; and (5) mechanically operated vibrator controlling shunt 
field. 

Differential Compound Generator. The differential compound 
generator is similar to an ordinary compound machine except that 
the series field is reversed. When the car attains sufficient speed 
(usually about 8 to 12 miles per hour) so that the generator 
voltage is high enough to charge the battery, the circuit is closed 
by a relay. Since at this speed the charging current is small, 
therefore the series field is weak and the flux is full value. If 
the generator speed is increased the voltage tends to rise, but 
this rise is limited, since the increased current develops an oppos- 
ing magnetomotive force in the differential series field and the 
flux of the generator is therefore reduced. The net result is that, 
as the speed increases, the charging current approaches constant 

* Winding 4 pole, two circuit, retrogressive 



126 



DESIGN OF SMALL MOTORS 



127 



value. This type of generator was used by the Delco Company 
on some of the Oakland, Olds, and Buick 1913 and 1914 models, 
on the Ahrens-Fox 1918 model, and others. In the preceding 
section it was shown how a small d.c. motor can be rewound to 
make such a generator. The advantage of this type is its extreme 
simplicity and the impossibility of its getting out of adjustment; 
its disadvantage is the fact that more field copper is required, 
resulting in a larger and more expensive generator for a given 
output. Since to adjust the charging rate is impossible, it 



yV2 450 -5RD. BRUSH REG. 
AUTOMOBILE GENERATOR HOT REGULATION 



% ^TAKEN IMMEDIATELY AFTER STANDARD TEMPERATURE TEST 




1000 1500 2000 

REVOLUTIONS PER MINUTE 

Fig. 41. High-Speed Performance Curyes of Automobile Generator 
Courtesy of Westinghouse Electric & Manufacturing Company 



3000 



may prove somewhat harder on the battery, as the latter is 
likely to be overcharged if the car is run at high speed for long 
periods. 

Third-Brush System. The third-brush system of voltage regu- 
lation is very extensively used. It takes advantage of the fact 
that the armature reaction of a generator twists the field flux in 
the direction of rotation. A third, or additional, brush is located 
between the main brushes. The shunt field of the generator is 
fed from the voltage between this third brush and the main brush 



127 



128 



DESIGN OF SMALL MOTORS 



behind it (with respect to direction of rotation). Thus, as the 
speed increases and the flux is distorted, the field voltage is 
reduced and the current output of the generator is limited. On 
further increase of speed the charging current is reduced, as 
shown in Fig. 41. Detailed information on one of the Westing- 
house Company's third-brush generators is given later. The gen- 
eral appearance of their line of generators is shown in Fig. 42. 
The third-brush type of generator was used in the 1918 models 
to a very large extent; for example, in the Buick, Cadillac, Cole, 
Hudson, Olds, Pierce Arrow, Packard, and many other cars. Fig. 43 




Fig. 42. Automobile Generators: (Top) with Ignition-Carrying Bracket; (Left) for 

Flange Mounting; (Right) for Cradle Mounting 

Courtesy of Westinghouse Electric & Manufacturing Company 

shows a Delco generator and ignition unit used on 9-N Continental 
engines. Fig. 44 shows the Auto-Lite Company's generator. 

Mercury- Tube Regulator. The other three types of voltage con- 
trol for generators all depend upon varying a resistance in series 
with a shunt-field circuit excited from the main brushes. A type 
known as the mercury-tube regulator was used by the Delco 
Company extensively on 1913, 1914, and 1915 models in the fol- 
lowing cars: Buick, Cadillac, Cole, Oakland, Stevens-Duryea, and 
others. In this regulator a plunger dips into a mercury well and 
the varying height of the mercury level short-circuits the requisite 
amount of a resistance which is in series with the shunt field. 
The plunger is controlled by a solenoid whose winding is across 



128 



DESIGN OF SMALL MOTORS 



129 



the generator terminals. This system has not proved entirely 
satisfactory and has not been used much in the last few years. 

Vibrating-Contact Regulator. Another regulating circuit which 
varies a resistance in series with a shunt field is the vibrating- 
contact type, similar in principle to the Tirrill regulator used in 
central station power equipment. It consists of a relay operating 
on narrow voltage variations; its winding is in series with the 
charging leads and its armature opens and closes a pair of con- 




Fig. 43. Delco Ignition Distributor and Generator as Used on 9-N Continental Motor 
Courtesy of Dayton Engineering Laboratories Company, Dayton, Ohio 

tacts which short-circuit a resistance in series with the shunt field. 
The action is as follows: If the car is running above the cut-in 
speed the current rises until the relay operates, opening the con- 
tacts and introducing resistance in the shunt field. The current 
then falls, owing to reduced field current, until the relay releases, 
closing the contacts again, after which the cycle is repeated. 
Actually the vibration is more or less continuous and is more 
rapid at higher car speeds. This system was used by the Delco, 
Westinghouse, and Bijur companies on 1915 and 1916 models, 
for example, on the Olds, Oakland, and Hudson cars. 

In 1918 . the Westinghouse Company put out a modified 
vibrating regulator which is claimed to be more reliable. It is 



129 



130 



DESIGN OF SMALL MOTORS 



I 



exactly the same as the regulator just described except that the 
back contact of the relay (which engages with the contact carried 
by the armature), instead of being fixed, is made to oscillate 
mechanically by a cam attached to the generator shaft. The relay 
magnet is thus relieved of the work of breaking the contact, and 
the regulation is claimed to be more sensitive and reliable. 

The Remy Electric Company introduced in 1916 a thermostat 
which is so adjusted that in winter it short-circuits a portion of 




GG-l A 






GG- 2 



GG-Z004 



GG-5 A 




Fig. 44. Auto-Lite Generator Parts as Used on Empire Cars 
Courtesy of Electric Auto-Lite Corporation, Toledo, Ohio 

the resistance in the shunt field of the generator and thus increases 
the charging rate. This compensates for the fact that not only 
is the drain on the battery greater during w^inter months when 
the engine is cold but the capacity of the battery is less at low 
temperatures. 

THIRD=BRUSH GENERATOR 
Design Data. At the present writing (1919) the third-brush 
generator is probably more widely used than any other regulating 



130 



DESIGN OF SMALL MOTORS 131 

circuit. The following design is an example of a third-brush gen- 
erator, frame No. 450, made by the Westinghouse Company. 
Fig. 42 shows the general appearance; Fig. 45 shows the circuit; 
and Fig. 41 gives the performance characteristics. It is seen that 
the generator cuts in (by an automatic potential relay) at about 
500 r.p.m. and may be operated up to 3000 r.p.m. The charging 
current reaches a maximum value at 1350 r.p.m., which cor- 
responds to about 25 miles per hour. At still higher speeds the 
current is less. This prevents overcharging the battery in case 
high speeds are maintained over long periods. As a matter of 
design it is instructive to note that the voltage between the 
third brush and negative, or ''field volts," is 69 per cent of the total 
voltage at the cut-in point; 45 per cent at the maximum current out- 
put; and only 12 per cent at 3000 r.p.m., or the maximum speed. 
This shows strikingly how the flux shifts to one side of the pole 
at high speeds. The power delivered by the generator is zero at 
the cut-in speed of 500 r.p.m.; 32.2 watts at 750 r.p.m.; 59.3 
watts at the point of maximum current at 1350 r.p.m.; and 21 
watts at 3000 r.p.m. 

The following tabulation gives the principal data of the design: 

Armature 

Diameter 3.135 inches 

Length (net) 1.937 inches 

Number of slots 21.0 

Slot pitch 0.4672 inch 

Slot opening 0.11 inch 

Slot depth 0.6925 inch 

Width of tooth at neck (uniform) 0.172 inch 

Thickness of punchings 0.017 inch 

Diameter of shaft 0.75 inch 

Number of commutator bars 41.0 

Diameter of commutator 1.75 inches 

*Number of inductors per slot 16.0 

Inductor section No. 16 double cotton-covered 

Field 

Outside diameter 5.25 inches 

Length 1.937 inches 

Number of poles 4.0 

Pole arc 1.5 inches 

Pole-core section 1.5 inches X 1.937 inches 

Yoke depth 0.375 inch 

Air gap 0.032 inch 

Field coil (all poles) 1)6.0 turns 

Field conductor No. 17 enamel-covered 

Field resistance (total) 1.75 ohms 

*Winding four pole, two circuit, singly re-entrant, with one "dead" armature cojl. 



181 



132 



DESIGN OF SMALL MOTORS 



The method by which these windings were computed is the 
same as that given in the designs of motors, with the exception, 
of course, that the IR drop in the armature must be added to 




the terminal voltage in order to obtain the internally generated 
e.m.f., which is analogous to the back-e.m.f. in a motor. 

One special feature of this design is that at maximum cur- 
rent at 1350 r.p.m.. Fig. 41, the field ampere turns per pole are 



132 



I 



DESIGN OF SMALL MOTORS 133 



186.8 while the armature ampere turns per pole are 328, making 
the ratio 0.57 instead of 1.25, which is used in motor design. 
<This large armature reaction is necessary in order to shift the field 
flux across the pole face and thus make possible the third-brush 
system of regulation. On account of the low voltage, commuta- 
tion difficulties are not encountered. 

K VIBRATOR=CONTROLLED GENERATOR 

Distinctive Feature. The design of a vibrator-controlled gen- 
erator does not differ materially from that of a third-brush gen- 
erator, except that the field has a series resistance in it which is 
short-circuited periodically by the vibrator and the entire field 
circuit is excited from the main generator leads. This resistance 
should be so proportioned that, if it is not short-circuited, it will 
reduce the field current sufficiently so that at maximum speed the 
generator voltage will have the desired value. Using the preced- 
ing generator as an example, if it were desired to convert it into 
a generator of the vibrator-controlled type, the field resistance 
should be about 15 ohms. This number is obtained from Fig. 41, 
which shows that the field voltage at maximum speed (3000 
r.p.m.) is 1 volt, which with a field resistance of 1.75 ohms gives a 

field current of — — , or 0.572 ampere. Thus, if the field were 
1./ o 

excited from the main generator terminals, the resistance of the 

circuit, in order to give 0.572 ampere with a voltage of 7, should be 

7 , 
— — — , or 12.23 ohms. Subtracting the field resistance of 1.75 

ohms, the additional resistance unit should be 10.48 ohms. It is 
well, however, to allow a margin in case the speed of the car 
exceeds the expected maximum, so that 15 ohms would be a good 
value. 

COMBINED STARTING MOTOR AND CHARGING GENERATOR 

Extent of Use. A combined starting motor and charging 
generator known as the one-unit system was employed by the 
Westinghouse Company on Hupmobile cars in 1915, and by the 
General Electric Company and the Gray & Davis Company on 
Ford cars. Fig. 46 shows the Gray & Davis motor-generator, and 



133 



134 



DESIGN OF SMALL MOTORS 



Fig. 47 the Delco used on Cadillac ears. The Bijur Company 
has employed a motor-generator on Scripps-Booth cars. Motor- 
generators are sometimes called ''genemotors/' a name used par- 
ticularly by the General Electric Company. 

Description. The Westinghouse Company's machine, which is 
typical, has a single armature winding and both series- and shunt- 
field windings. At starting, a heavy current flows through the 
series winding, and this supplies a strong field, which gives the 
necessary torque. When the engine accelerates, the current taken 



stop Collar 
Screw Shaft 
pinion 



^Hdjusling Screw Jf- Lock Nut -S 

■ Clamping Nuls - 7 



Oil4- 

Motor Terminal-1 




Dijnamo 
Sprocket 



3 

Di^namc Qea. 



5-Clampinq Huts 
Hdjusting Bracket 

Mounting S racket 



Fig. 46. Details of Gray & Davis Generator Unit for Far i Starter 

from the battery rapidly decreases to zero and then reverses in 
direction, thus charging the battery. A shunt field is provided, which 
is fed from a third brush and gives voltage regulation in the same 
way as in the previously described third-brush generator. When 
charging, the current is small compared to the starting current, 
so that the series winding develops only a small magnetomotive 
force. The complete machine weighs 55 pounds and is suitable for 
cars up to 200 cubic inches cylinder displacement. The most 
favorable field of application for the combined starter and gen- 
erator is on small cars having engines of small cylinder displace- 



134 



DESIGN OF SMALL MOTORS 



135 



ment. It is then possible to keep down the weight to a reason- 
able value and secure the advantage of simplicity. On large cars 
with engines of greater cylinder displacement the two-unit system 
is more economical as regards both weight and cost. 




Fig. 47. Delco Motor-Generator for Cadillac Automobiles 
Courtesy of Dayton Engineering Laboratories Company, Dayton, Ohio 

FARM LIGHTING OUTFITS 

Description. Farm lighting outfits, which are now widely 
used, consist of a storage battery, a charging generator, and a 
source of power, usually a gasoline or kerosene engine, to drive 
the generator. The battery is permanently connected to the sup- 
ply mains of the lighting circuit, and thus light is available at any 
time of day or night. The battery requires charging from one to 
three times a week under average conditions. The more recent 
outfits have been designed so that they may also supply power, 
either electrically to motors connected to the lighting circuit or 
mechanically from the engine. 

Voltage Regulation. An essential requirement of farm light- 
ing outfits is proper voltage regulation, both as to momentary 



135 



136 



DESIGN OF SMALL MOTORS 



fluctuations due to the explosions of the engine and the range of 
variation dependent upon the state of the battery charge. A 
high voltage shortens the life of the lamp filaments, while a low 
minimum voltage reduces the candle-power of the lights. In early 
forms the battery was disconnected from the lighting circuit 
during charging, thus avoiding the rise in voltage during that 
time and also the flickering due to engine pulsations. This caused 
inconvenience, as frequently it would not be realized until after 




Fig. 48. Farm Lighting Outfit, Automatic-Regulator Type 
Courtesy of Western Electric Company, New York 



dark that the battery required charging, and it would then be 
necessary to do without light at the very time it was needed. 
The present practice is to have the battery connected to the 
lighting circuit at all times and to provide suitable means of volt- 
age regulation. 

Belt=Driven Systems. As gasoline engines are frequently 
employed for other purposes, one form of outfit employs a belt- 
driven generator. In case the farmer already has an engine for 
other work the cost of the lighting equipment is thus reduced. 



136 



DESIGN OF SMALL MOTORS 



137 



while if it is necessary to purchase an engine it may be used for 
other purposes and may be mounted on a portable base. 

Automatic- Regulator Type. In Fig. 48 is shown a belt-driven 
outfit made by the Western Electric Company. It is known as 



CONTACTS 



GENERATOR 
FlEUO 



CUTOUT 
REUAY 



COMTACT^ 



COMTROL 

MAGHET 

RESISTANCE 



itLECTRK: GAUGE 
CINDtCATtHG CHARG& 
^AMP OiSCHARGE) 




STORAGE. 
BATTERY 



REQULATIH^ 
FIELD 



GENERATOR 



UOAD 



Fig. 49. Wiring Diagram of R-7 Automatic Regulator 
Courtesy of Western Electric Company, New York 

the automatic-regulator type and is sold in sizes of 500-1000 
watts generator capacity with sixteen-cell batteries of 50-180 
ampere hours capacity. Fig. 49 shows the circuit. 

When starting, the cycle of operations is as follows: The 
switch is thrown to the side marked START, and current flows 



137 



138 



DESIGN OF SMALL MOTORS 



from the positive battery terminal to the generator, which accel- 
erates as a motor, thus cranking the engine. When the engine 
begins to operate by its own power, the switch is thrown to the 
position marked RUN, When the voltage of the generator 
becomes sufficient to charge the battery, the cut-out relay recon- 
nects the generator to the battery. The voltage continues to 
rise until the vibrating regulator shown as Control Magjiet begins 




-».^:y'V*.vA:^t ^ 



Fig. 50. Type F Lighting Plant with Closed Water-Jacket Engine 
Courtesy of Fairbanks, Morse & Company, Chicago, Illinois 



to operate. When its armature is pulled down, resistance is intro- 
duced into the regulating field circuit, which weakens the flux and 
causes the voltage to drop until the armature releases again and 
the cycle recommences. By this means the voltage is held 
constant, giving a gradually tapering charge to the battery. 

The generator itself is an ordinary shunt machine with the 
addition of the regulating field. It could be designed by the same 
rules given in the designs of d.c. motors. 



138 



1 



DESIGN OF SMALL MOTORS 



139 



Hand-Regulated Type. A similar belted outfit made by 
Fairbanks, Morse & Company is shown in Fig. 50. Hand regulation 
is provided in this outfit by means of a rheostat in the shunt 
field of the generator. The engine is designed to run on kerosene 



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and is furnished in 3-horsepower and 6-horsepower sizes, giving a 
corresponding generator output for 32 and 144 lights of 20 watts 
each at 32 volts. The batteries range from a capacity giving 19 
lights for 5 hours followed by 13 lights for 8 hours to a capacity 
giving 56 lights for 5 hours followed by 40 lights for 8 hours. 



139 



140 



DESIGN OF SMALL MOTORS 



OUTUNE, 



CFBfiSE 



£LECrf?fC G/7i/GE /VO/C/tT/NO 
"C H/fRGE. "OFF"a^DfSCH/1R G E" 




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70 fCN/r/ON CO/L 



Fig. 52. Wiring Diagram for 32-Volt Direct-Connected Lighting Outfit 
Courtesy of Western Electric Company, New York 



140 



DESIGN OF SMALL MOTORS 141 

Direct=Connected Sets. In Fig. 51 is shown the Western 
Electric 1.5-kilowatt 32-volt set direct connected to a 2.5-horsepower 
engine, which may be operated on either gasohne or kerosene. 
Fig. 52 shows the circuit connections. When the starting-switch 
handle is pressed down, current flows from the positive battery 
terminal to the control panel and to contact 5. Here the current 
divides, the main portion flowing through contact 6 to the series 
field of the generator. Flowing through the series field, the cur- 
rent enters the armature through the positive brush and leaves by 
the negative brush, flowing back to the control panel, through the 
upper knife switch C and the lower fuse 6r, and back to the storage 
battery through its negative terminal. It will be noted that the 
shunt field receives its current from the positive side of the gen- 
erator within the generator frame and from the center clip of 
switch C, which is negative. The armature begins to revolve 
when current is applied in the manner mentioned and operates 
the engine rapidly. With the starting-switch lever held down the 
engine will continue to rotate, the generator acting as a driving 
motor and receiving current from the storage battery. 

The engine then begins to fire and to operate by its own 
power, the speed rapidly increasing to the normal running point. 
The engine ignition circuit receives current from the storage bat- 
tery in starting through the starting-switch contact 7 and switch 
clip 9 of the upper switch C. When the engine begins to operate 
by its own power, the generator voltage rises very rapidly. This 
increases the excitation of the shunt, or inner, winding of the 
relay magnet A, which was connected by the switch contact 8 
when the starting switch was pressed down. At the proper gen- 
erator voltage this magnet excitation is sufficient to close contacts 
H and 15 by attracting the armature lever 16, The closing of 
the main contacts H connects the positive brush of the generator 
with the storage battery through the relay-magnet series winding 
17 and the electric indicator ¥, When this takes place the start- 
ing-switch handle may be released, which opens the rear contacts 
5, 6, 7, and 8. The battery charging current flowing through the 
series winding is sufficient to hold contacts 14 and 15 closed with- 
out the aid of the shunt, or inner, winding, which was disconnected 
when the starting-switch contacts 8 opened. At this stage the 



t41 



142 DESIGN OF SMALL MOTORS 

current for engine ignition is supplied through the relay-magnet 
contacts 15. The current flowing to the storage battery through 
the electric indicator causes it to indicate CHARGE, which 
shows that the storage battery is being charged. 

As the charge continues, the current flowing through the bat- 
tery gradually grows less, and the battery voltage and specific 
gravity rise. As the current decreases, the effective magnetic 
pull on armature 16 also decreases until a point is reached, when 
the storage battery is sufficiently charged, at which the control 
spring 18 overpowers the magnetic pull and the armature springs 
back, opening contacts H and 15, which action stops the engine 
by opening the ignition circuit through contacts 15 and discon- 
nects the storage battery through contacts H. 

The generator is a four-pole machine with a two-circuit 
winding, thus requiring only two sets of brushes, which are acces- 
sible upon removing a cover plate which protects the commutator. 
The storage battery is furnished in two sizes which have capaci- 
ties of 90 and 180 ampere hours at 32 volts, corresponding 
respectively to 2700 and 5400 watt hours. 

The Delco Company makes a direct-connected outfit in which 
a distinguishing feature is the system of voltage regulation. A 
magnet is provided with a potential winding which is connected 
across the generator terminals. The armature of the magnet 
operates a butterfly valve, which controls the amount of explosive 
mixture admitted to the engine, so that if the voltage tends to 
fall the speed of the set is increased, and vice versa. 

Size of Outfit. The following information is useful in esti- 
mating the size of battery required for any proposed installation: 

The average lighting load per day for a farm is approximately as follows: 

1 20-watt lamp in kitchen 3 hours per day = 60 watt hours 

1 20-watt lamp in dining room 2 hours per day = 40 watt hours 

1 40- watt lamp in hving room 3 hours per day =120 watt hours 

1 20-watt lamp in each of 3 bedrooms | hour per day = 30 watt hours 
3 20-watt lamps in barns 2 hours per day =120 watt hours 

Miscellaneous, such as porch, yard, and basement = 30 watt hours 

Total average lighting load per day = 400 watt hours 

Dividing the capacity of the battery expressed in watt hours by this average 
daily load will give the number of days that the battery will run on a ful] 
charge. In the case of a battery with a capacity of 2700 watt hours, dividing 



142 



DESIGN OF SMALL MOTORS 143 

by 400 (average light load per day) will give 6f as the number of days that the 
battery will operate on a full charge. 

The average electric iron consumes 500 watts per hour. With 2700 watt- 
hour capacity, dividing by 500 will give 5|, which is the number of hours the 
battery will operate an iron on a full charge. 

The average j-hp. motor uses about 300 watts per hour. With a 2700 
watt-hour capacity, dividing by 300 will give 9, which is the number of hours 
the battery will operate such a motor on a full charge. 

The size of the generator should be such that between four 
and eight hours is required to charge the battery. After deciding 
the output of the generator the required horsepower of the engine 
may be computed, making allowance for the efficiency of the 
generator. An efficient outfit will charge a 50 ampere-hour 32-volt 
battery on from 1 to 1.5 gallons of gasoline. 



143 




TRANSFORMERS ON PLATFORM AT PLANT OF GEORGE W. JACKSON 
COMPANY, CHICAGO 



DESIGN OF 
SMALL LOW- AND HIGH- 
TENSION TRANSFORMERS 



INTRODUCTION 

Use of Transformers. The ordinary transformer is a very 
familiar object and plays a very important role in our everyday 
existence. While in construction it is a very simple device and in 
operation a very powerful but withal a very quiet device, the 
theory of its operation is somewhat complex. 

Increase of Pressure. The transformer is used in the trans- 
mission and distribution of power to save copper in line construc- 
tion. Electric power, being the product of electric pressure and 
electric current, may result from multiplying a very large current 
by a very small pressure or from multiplying a very small current 
by a very large pressure; that is, in transmitting any given power, 
if the pressure factor be increased, the current factor may be 
decreased, and the current factor determines the cross-section, or 
size, of copper line conductors necessary. 

At the present time very large amounts of power must be trans- 
mitted very considerable distances, and saving in capital invested in 
copper is imperative in order that power enterprises shall meet with 
financial success. It has been found that the amount of copper 
necessary for transmitting a given amount of power a given distance 
depends inversely upon the square of the pressure; that is, if the pres- 
sure between transmission lines is doubled, only one-fourth as much 
copper is necessary to transmit the same power over the same distance 
with the same line loss. If the pressure is increased three times, 
the necessary amount of line copper is only one-ninth. 

In the transmission of electrical power the function of the 
transformer is to raise, or ''step up," the electrical pressure from a 
value that is found possible at the generator to such a higher 
value as to save copper in large amounts, reducing the amount of 
fixed capital tied up and effecting a continuous saving in interest 
and investment. It is obvious that the interest on the cost of the 



145 



2 DESIGN OF SMALL TRANSFORMERS 

transformer plus the cost of its operation must be less than the 
interest charge on the amount of copper it displaces. 

Alternating-current generators, as at present constructed, pro- 
duce pressures of about 10,000 volts, so that transformers for trans- 
mission service must ^'step up" the 10,000 volts to as high as 200,000 
volts to meet the requirements of the larger power transmissions. 

Reduction of Pressure. While it is desirable to resort to very 
high pressures for transmission purposes, such pressures are unde- 
sirable when electrical power is brought inside buildings, as high 
pressures are dangerous to human life and introduce a fire risk so 
far as buildings and contents are concerned. It is therefore neces- 
sary to reduce the high pressures of transmission when power is 
distributed among buildings or within densely populated areas. 
The transformer is again used, this time to reduce, or ''step 
down," high pressures to a desirable low^ pressure. 

Definition of Transformer. It will be well at this point to 
have a working definition of a transformer. A transformer is a 
device to which electrical power is supplied at a definite pressure 
and from which electrical power is obtained at a different pressure. 
The transformer merely changes one factor of power, or energy. 
It does not change one form of power into another, as does a 
generator or a motor. The fact that one factor of power, the 
pressure, may be increased does not mean that something is 
being created by the transformer. Every increase in pressure 
means a corresponding decrease in current. 

Having defined a transformer with reference to its function, 
its mechanical construction should next be considered. A trans- 
former consists of two electrical circuits interlinked by a magnetic 
circuit. This should appeal to Odd Fellows who recognize the 
insignia of the "three links." One of the two electrical circuits is 
designated as the 'primary, and the other as the secondary, while 
the link connecting the two is the core. 

THEORY OF TRANSFORMER OPERATION 

ELEMENTARY PRINCIPLES 
Relation between Electric Current and Magnetic Field. It 

was first discovered about 1819 that when an electric current was 
allowed to flow in a conducting wire of copper, a magnetic field 



146 



DESIGN OF SMALL TRANSFORMERS 3 

always existed in the space around the conductor. This was the 
beginning of electrical engineering, and in the consideration of 
transformers it is very important to have a very thorough under- 
standing of the fixed relation existing between the direction of the 
current in a conductor and the magnitude and the direction of the 
resulting magnetic field. This relation is shown graphically in 
Fig. L A circular conductor ABC is connected with a battery of 
three cells E connected together in series and sending a current 
around the circular conductor in a counterclockwise direction, as 
denoted by the large arrow near C. The magnetic lines of force 
always surround current-carrying conductors in the form of com- 




Fig. 1. Magnetic Lines of Force around Conductor 

plete curves, which are circular in the case of a straight wire 
removed some distance from other current-carrying wires. In the 
illustration the surrounding magnetic lines are shown in circles 
having directions indicated by the arrowheads. 

Right- Hand Screw Rule. The relation between the current 
and the magnetic field is further emphasized in Fig. 2. The 
inner circle denotes the cross-section of a conductor arranged per- 
pendicular to the page, in which the current is flowing down, or 
away from the observer. The magnetic field produced by the 
current consists of circular magnetic lines surrounding the con- 
ductor and having a clockwise direction. This leads to what 
might be called the right-hand screw rule, stated as follows: The 



147 



4 DESIGN OF SMALL TRANSFORMERS 

direction of advance of a right-hand screw corresponds with the 
direction of current, ivhile the direction in which the screw is turned 
corresponds with the direction of the magnetic field surrounding the wire. 
Another phase of the relation is the fact that a current flow- 
ing around a coil in a counterclockwise direction produces a north 
magnetic pole. In Fig. 1 the current around the single turn of 
conductor is counterclockwise and the magnetic lines of force are 
coming up through the loop, or turn. A north, or N, pole is one 
from ivhich magnetic lines are pointing, while a south, or S, pole is 
one into ivhich magnetic lines are pointing. 



cVcjzJ^:ii^o^ 




T 



Fig. 2. Direction of Magnetic Lines of Force 
with Relation to Current 

Magnetic Lines of Force. The idea of magnetic lines of force 
is a very helpful one, and the direction of the lines may be 
ascertained by employing a small compass needle, the N end of 
which will always point in the direction of the lines. The presence 
of magnetic lines in any circuit or device may be made evident by 
holding a large cardboard sheet over the device and sprinkling 
fine iron filings over the surface of the card. 

Magnetic Conductivity, or Permeability. It is very important 
to note that different media allow magnetic lines of force to pass 



148 



DESIGN OF SMALL TRANSFORMERS 5 

through them with different degrees of ease. For example, air is 
not a very good conductor for magnetic Hues, neither is copper nor 
zinc. Iron, however, is an extremely good conductor for magnetic 
fields, or lines of force, and is therefore employed in place of air or 
other substances whenever possible. The advantages of employing 
iron instead of air as a medium for conducting magnetic lines may 
be appreciated better when it is stated that from experiment, with 
a given coil of wire consisting of a definite number of turns in 
which a certain current flows, if iron replaces the air inside the 
coil, the number of magnetic lines of force threading through the 
coil may be increased over 2000 times. This is the same as 
assuming that the single turn shown in Fig. 1 with a certain cur- 
rent in the loop might produce 200 lines of force with no iron near 
the arrangement, and that, without in any way changing the 
amount of current or the electrical circuit, if a soft-iron core were 
introduced so as to pass through the circular loop, the number of 
magnetic lines might be increased to 200X2000, or 400,000 lines. 
The advantage of iron as a magnetic conductor is very evident. 

The measure of the ease with which certain substances con- 
duct magnetic lines of force is called magnetic permeability, or 
simply permeability. The Greek letter /x is often used to denote 
permeability. While iron is found to be the best conductor of 
magnetic fines of force, the various grades or kinds of iron differ 
greatly in magnetic conducting ability. The softer the iron the 
greater its conductivity, or its permeability; therefore it is advis- 
able to employ pure soft iron in the cores of such devices as 
transformers. 

Retentivity. One important characteristic of iron needs con- 
sideration in connection with its use in magnetic circuits, namely, 
retentivity. Whenever the current is shut off from such a circuit 
as shown in Fig. 1, the magnetic lines all disappear or fade away. 
Should there be an iron core in the loop when the current is 
stopped, a considerable number of magnetic lines will remain in 
the iron, and the harder the iron the greater will be the number 
of the retained lines. If very hard steel is used as a core instead 
of soft iron, many more lines will be retained. The steel will 
have become a so-called permanent magnet under such conditions. 
The characteristic of retaining magnetic lines after the magnetizing 



149 



6 



DESIGN OF SMALL TRANSFORMERS 



cause has been removed is termed retentivity. Since high values 
of retentivity are very undesirable in iron used for transformers, 
such iron should be very soft. 

Magnetic Induction. Whenever lines of force are produced, or 
set up, by the electric current, whether in air, steel, or iron, the 
magnetization is called magnetic induction or simply induction. 
When the current in a circular loop. Fig. 1, is 1 ampere, a certain 
number of magnetic lines are set up in the air magnetic circuit. 
Since the unit for electric current is not of the same kind as the 
unit for magnetic lines, the relation between the number of 
amperes of current and the number of lines must be expressed 
symbolically. In order to produce a given amount of magnetic 
field, or number of lines, a current of a certain number of amperes 
must exist in the circular loop. 

ELECTROMOTIVE FORCE OF INDUCTION 

Induced Current. In the foregoing paragraphs there has been 
some discussion of magnetic induction, that is, the setting up of 



/f 




Fig. 3. Current Induced by Magnetic Lines of Force 



magnetic lines of force, or so-called magnetic flux, by an electric 
current. The reverse of this relation is of equal importance; 
namely, given the proper conditions, a changing magnetic field 
will produce an electric current. The relations of a changing 
magnetic field and a resulting induced current may best be 
explained by referring to Fig. 3. Let AB denote a conductor 
that may be moved back and forth along the rectangular circuit 



150 



aenol 



DESIGN OF SMALL TRANSFORMERS 



ehoted by CD EG through the lines of force of a magnetic field 
having a direction downward, as indicated by the arrows. Sup- 
pose the slider is moved to the right, as indicated by the arrow 
M; according to the right-hand rule, which is given in the follow- 
ing paragraph, an electric pressure having a direction from A to 
B will be induced in the moving slider AB. 

■ Right=Hand Rule. The so-called right-hand rule should be 
very thoroughly understood by any one who wishes to become 
proficient in the construction or operation of transformers. The 
rule is as follows: Hold the right hand so that the thumb and the 
forefinger are perpendicular to each other and so that the second 
finger is perpendicular to both the thumb and the forefinger. Next 




Fig. 4. Graphic Representation of Right-Hand Rule for Induced Currents 

place the hand in a magnetic field so that the forefinger points in the 
direction of the magnetic lines and the thumb points in the direction 
in which an inductor is moved across the lines of force. The second 
finger will then point in the direction of the induced pressure. 

This rule will become evident if Fig. 4 is studied carefully. The 
magnetic lines issue from the N pole as shown by the forefinger. 
The lines are produced by a current from a battery of four cells 
E passing around the pole in a counterclockwise direction. If the 
moving inductor AB cuts the lines in the direction indicated by 
the thumb, the induced pressure will be from A to B, as indi- 
cated by the second finger. 

Results of Moving Slider to Right. Returning to a considera- 
tion of Fig. 3, it should now be observed that an induced pressure 



151 



8 DESIGN OF SMALL TRANSFORMERS 

from A to B m the moving inductor will cause a current to flow 
around the closed circuit ABBE in a counterclockwise direction. 
Further observation will show that a motion of AB to the right 
in effect causes more and more lines to be included by, or intro- 
duced into, the closed circuit ABDE. The next logical step in 
the process of our study is to obtain some information as to 
the amount of pressure induced and the amount of the resulting 
current. 

Fundamental Equation of Induced Pressure. The relation 
between the strength of the magnetic field, the velocity of the 
moving inductor AB, and the amount of the induced pressure is a 
very simple one and may be expressed in symbols as follows: 

in which E' denotes the induced pressure in volts; / denotes the 
length of the moving inductor in centimeters (see Table I for rela- 
tion between an inch and a centimeter); ^ denotes the density of 
the magnetic field in gausses, that is, in number of lines of force 
per square centimeter; and v denotes the velocity of the moving 
inductor in centimeters per second. A completely worked-out 
example will serve to illustrate this relation. 

Exam-pie. A wire 30.48 centimeters long moves with a velocity of 2682.24 
centimeters per second across a magnetic field in which there are 1000 lines of 
force per square centimeter, or 1000 gausses. Find the value, in volts, of the 
pressure induced between the ends of the moving inductor. 

The fundamental equation applying to this case is the one just given, 

108 

Since I equals 30.48 centimeters, ^ equals 1000 gausses, v equals 2682.24 centi- 
meters per second, and 10^ means 10 raised to the eighth power, that is, multi- 
phed by itseM eight times, making the proper numerical substitutions in the 
equation will give 

„, 30.48X1000X2682.24 81754675.2 ^ ^, ,^ 

E = = =0.81 volt 

100 000 000 100 000 000 

Ans. 0,81 volt 

If the velocity v of the moving inductor AB, Fig. 3, is 2682.24 centimeters 
per second and the active length I of the inductor is 30.48 centimeters, then the 
area A swept over by the moving inductor during one second must be 30.48 X 
2682.24, or 81754.6752 square centimeters. If each square centimeter of this 
area contains 1000 hues of force, then 1000X81754.6752, or 81754675.2 lines of 
force must be introduced into the circuit represented by ABDE in 1 second. 



152 



1 



DESIGN OF SMALL TRANSFORMERS 9 

TABLE I 
Table of Equivalents 

LENGTHS, AREAS, AND VOLUMES 

1 inch =2.54 centimeters 

1 centimeter =-^=0.393 inch 
2.54 

1 square inch =2.54^ = 6.45 square centimeters 

1 square centimeter = =0.155 square inch 

6.45 

1 cubic inch =2.543 = 16.38 cubic centimeters 

1 cubic centimeter = =0.06105 cubic inch 

16.38 

1 mil = inch 

1000 
1 circular mil =area of a circle 1 mil in diameter 

= area of a circle inch in diameter 

1 square mil =area of a square 1 mil on a side 

A square mil is greater than a circular mil, since the area of a square is 

greater than the area of an inscribed circle. 

1 square inch = 1000 X 1000 = 1 000000 square mils 

. , 4X1000 000 , ^-. _^^ . , ., 
1 square mch = = 1 274 500 circular mils 

WORK AND POWER 

1 foot pound =unit of work, or work required to raise 1 pound 1 foot 

1 British Thermal Heat Unit = 778 foot pounds 

* . The ordinary symbol for this unit is B. t. u." One B. t. u. may be defined 
as the amount of energy, or work, necessary to raise the temperature of 1 
pound of water 1° F., when the water is at its maximum density. 

1 horsepower = 33000 foot pounds per minute 

=- =550 foot pounds per second 

60 ' "^ 

= 746 watts = 373 volts X 2 amperes 

= 746 volts X 1 ampere 

= 2 volts X 373 amperes 

= 74.6 volts X 10 amperes, etc. 

Modified Equation of Induced Pressure. The number of lines 
of force in a unit area multiplied by the number of units in any 
given area must give the total number of magnetic lines in that 
area. That is, if ?^ denotes the number of lines per square centi- 
meter, A denotes the number of square centimeters in the given 
area, and $ denotes the total number of lines, or the total mag- 
netic flux, then 

^=^A 



153 



10 



DESIGN OF SMALL TRANSFORMERS 



It is evident from the foregoing explanation that the induced 
pressure in any circuit may be denoted by 



E' = 



tw 



volts 



in which t equals the time in seconds. This form of the funda- 
mental equation for induced pressure is employed very frequently 
in connection with transformer theory and design. 

Relative Motion between Circuit and Magnetic Field Neces= 
sary. It should be very carefully noted that an electrical pressure 
can only be induced while there is a relative motion between a wire, 
or a circuit, and a magnetic field. If the magnetic field that 
threads through a circuit is steady in value and the circuit is 

B 




Fig. 5. Effect of Broken Circuit on Induced Current 

stationary, then there can be no induced pressure. A direct cur- 
rent that does not vary in its amount from one second to the 
next cannot therefore be employed in cohnection with trans- 
formers to effect changes, or transformations, in pressures. It 
should also be noted that if the circuit through which the number 
of magnetic lines is changing is an open one, no current can flow 
in the circuit, although there is an induced pressure. This is 
illustrated in Fig. 5, which shows the same arrangement as does 
Fig. 3 except that there is a cut at tt' in the portion of the circuit 
between Z) and E. In such a case when AB i^ moved to the 
right, a pressure is induced as previously described and computed. 
The pressure exists between the portions of the circuit t and t' 



154 



DESIGN OF SMALL TRANSFORMERS 



11 



and according to the right-hand rule f will be the positive, or +,' 
terminal and t will be the negative, or — , terminal. If while AB 
is moving and the pressure is being induced, t and t' are brought 
together, thus closing the circuit, the induced current can flow. 

Left=Hand Rule. The opposite of the right-hand rule is quite 
naturally a left-hand rule and in transformer design and calcula- 
tion it is quite as important as the right-hand rule. The left-hand 
rule serves to bring out the relation between the current in a 
wire and the field existing in the space about the wire and is as 
follows : Hold the left hand so that the thumb, the forefinger, and the 
second finger are perpendicular to 
one another. Next "place the hand 
in a magnetic field so that the 
forefinger points in the direction 
of the lines of force and the sec- 
ond finger points in the direction 
of the current in a conducting wire 
placed in the field perpendicular 
to the lines. Then the thumb will 
point in the direction in ichich 
the wire will tend to move. 

This rule may be better un- 
derstood by referring to Fig. 6. 
The movable conductor is de- 
noted hy AB; the direction of the 
current is denoted by the arrows; and the direction of the mag- 
netic lines of force is from the N end of the magnet. The con- 
ductor AB will tend to move to the left. In Fig. 7 AB denotes a 
conductor that can slide along the parallel conductors CD and 
FG; the uniform magnetic field has a direction downward, as 
shown by the vertical arrows; and a battery of cells at E can 
send a current through the circuit DBAF in a counterclockwise 
direction. Under such conditions the sliding conductor AB will 
tend to move to the left as indicated by the arrow M, and accord- 
ing to the left-hand rule this motion removes lines of force from 
the circuit. 

Counter=Electromotive Force. According to the right-hand 
rule applying to an inductor moving across a magnetic field, 




Fig. 6 



Graphic Representation of Left- 
Hand Rulg for Induced Currents 



155 



12 



DESIGN OF SMALL TRANSFORMERS 



whenever the number of lines threading through a circuit is 
decreased, the induced pressure is clockwise around such an 
arrangement as represented by Fig. 7. This is just the reverse of 
the condition shown in Fig. 3. If, then, the current in the mov- 
able conductor AB, Fig. 7, is in such a direction and of such an 
amount as to cause AB to actually move to the left, thus reducing 
the number of lines of force, a pressure will simultaneously be 
induced in the circuit, according to the right-hand rule, in a clock- 
wise direction, or directly opposite to that of the applied pressure 
of battery E. 

This clockwise induced pressure due to the motion oi AB 
sends a current around the circuit in a clockwise direction. The 




^rtce ^" 



Fig. 7. Graphic Representation of Effect of Motion on Induced Currents 

resultant current in the circuit must therefore be the difference 
between the current produced by the battery and that resulting 
from the induced pressure. Whenever the conductor AB stops 
moving, there is no induced pressure and hence no induced current, 
and the current in the circuit is simply one whose value may be 
expressed by Ohm's law 

E 



7= 



R 



in which / denotes the current, in amperes, in the circuit, E 
denotes the pressure, in volts, applied to the circuit, and R 
denotes the resistance; in ohms, of the complete circuit. When- 
ever the conductor AB is moving to the left, the current in the 



156 



DESIGN OF SMALL TRANSFORMERS 13 

circuit is less than when the conductor is stationary. When the 
conductor is in motion, the current may be expressed by 

R 

in which F denotes the current in the circuit, E the appHed pres- 
sure, and R the total resistance as before, but e denotes the 
induced pressure which, according to the right-hand rule, is 
directly opposite in direction to the applied battery pressure. 
Because of its opposite direction this pressure is called the counter- 
electromotive force; and, being produced by induction or relative 
motion between an inductor and a magnetic field, it is called the 
electromotive force of induction. 

If the direction of the current around the circuit ABDEF is 
reversed, the movable portion AB will move to the right, accord- 
ing to the left-hand rule. In this case the number of lines of 
force threading through the circuit will be increased ii AB moves, 
and the resulting induced pressure will be counterclockwise, 
according to the right-hand rule, or again in a direction exactly 
opposite to that of the applied pressure from the battery. Intro- 
ducing lines of force into a circuit will produce an induced pressure 
equal in value to, but opposite in direction to, that produced by 
removing the same number of lines in the same time, the direction 
of the lines being the same in both cases. 

Rate of Change in Magnetic Flux. If the conductor AB, 

Fig. 7, moves to the left at a certain definite rate, the number of 

lines threading through the circuit is reduced in exact proportion 

to that rate. If the distance moved by the inductor is, say, 3048 

centimeters, or 100 feet, during 2 seconds, its velocity may be 

expressed as 

d 3048 , .^, 
^) = — =— - — = 1524 cm. per sec. 

t Zi 

or 

100 rr.,^ 

i;=-— = 50 it. per sec. 

If the length (A AB happens to be, say, 60.96 centimeters, or 2 
feet, and it has a velocity of 1524 centimeters, or 50 feet, per 
second, it will sweep over an area equal to 1524X60.96, or 
92903.04 square centimeters, vdiich equals 100 square feet. If ^, 



157 



14 DESIGN OF SMALL TRANSFORMERS 

or the number of magnetic lines per square centimeter passing 
downward through the circuit, happens to be 800, then 92 903.04 
X800, or 74,322,432 hnes, will have been traversed by the moving 
inductor during 1 second of its motion. 

Since the total area swept over by the moving portion of any 
circuit such as that in Fig. 7 multiplied by the magnetic density, 
or the number of lines of force per unit area, will give the number 
of Hnes introduced into or removed from the circuit, the induced 
pressure in any circuit due to the rate of change in magnetic lines 
included by it may be expressed by 

E' = -^ volts 

no^ 

The change in flux, or in the number of lines of force, is denoted 
by $, and the time during which the change takes place by t. 

Exam-pies. 1. If the magnetic flux, or the number of Hnes threading 
through a circuit, is changed from 10,000,000 to 1000 hnes in — second, find 
the resulting induced pressure. $ equals 10 000 000 - 1000, or 9 999 000 Hnes, 

and t equals second; substituting gives 

9 999 000 ^ 999 900 000 

"JLvlOs 100000000 
100^^^ 

= 9.99 volts 

Ans. 9.99 volts 

2. If the magnetic flux threading through a circuit changes from 100,001,000 
hnes to 1000 in — — second, find the resulting induced pressure. 

Ans. 100 volts 

3. If the number of magnetic lines passing through or included by a cir- 
cuit is changed from 1,000,001,000 to 1000 in — rp: second, find the resulting 

100 

induced pressure. 

Ans. 1000 volts 

The equation denoting the value of induced pressure due to 

the rate of change in magnetic flux may also be expressed by 

j^, $l-$2 u 

in which $i denotes the initial value of the number of lines and 
$2 denotes the final number of lines threading a circuit in a time 
denoted by t 



158 



DESIGN OF SMALL TRANSFORMERS 



15 



Rate of Change in Current. It is very important now to 
realize that if the inductor AB, Fig. 7, should remain stationary 
and if 74,322,432 lines should be uniformly removed each second 
from threading through the circuit, the pressure induced in the 
circuit would be exactly the same as though AB were moved to 
the left at the rate of 1524 centimeters per second, if its length 
between the portions of the circuit upon which it slides were 60.96 
centimeters and if the magnetic density were 800. Induced pres- 
sure always depends upon the relative motion of any circuit and a 
magnetic field threading through the circuit. The question may be 
raised as to how the number of lines threading through the circuit 




Fig. 8. Effect of Change in Current on Magnetic Lines of Force 

may be removed. One method may obviously be by reducing the 
amount of the current producing the field. 

Suppose that a variable resistance, or rheostat, is connected 
into the circuit as indicated in Fig. 8 at R. By moving a variable 
contact represented by J to the right, the resistance at Rx is 
increased, the current in the circuit is decreased, and therefore the 
number of lines is reduced. The quicker the contact is moved, 
the greater will be the rate of change in the current and the 
greater the rate of change in the number of lines. Should the cir- 
cuit be suddenly opened, thus suddenly stopping the flow of the 
current, the number of lines would be reduced to zero wry sud- 
denly — provided there were no iron in or near the circuit. 



159 



16 



DESIGN OF SMALL TRANSFORMERS 



Coefficient of Inductance. The counter-electromotive force 
produced according to the foregoing conditions depends in value 
upon the rate of change in the current in any circuit. Were pres- 
sure and current expressed in the same units, it might be said 
that the induced pressure would be equal to the rate of change 
in the current. Since, however, the units are unlike, the relation 
between the rate of change in the current in a circuit and the 




Fig. 9. Parts of Simple Transformer 

resulting induced pressure may be expressed by the following 
equation : 

t 
that is, the pressure in volts E' induced in any circuit by a rate 
of change in the current of I amperes in any time t will be L 
times the rate of change in the current per second. L therefore 



160 



DESIGN OF SMALL TRANSFORMERS 



17 



simply represents a coefficient, or a multiplier, by which the 
numerical value of the rate of change in the current in a circuit 
per second must be multiplied to give the numerical value of the 
induced pressure in volts. L is called the coefficient of induction 
and is rated in a unit called the henry. 

Unit of Inductance. The significance of the unit of induc- 
tance, the henry, may perhaps be brought out by considering its 




Fig. 10. Effect of Opening Battery Circuit of Simple Transformer 

numerical relation. Suppose the current in a circuit should 
change at the rate of 1 ampere per second. If the coefficient of 
inductance of the circuit should be 1 henry, then the induced 
pressure would be exactly 1 volt. 

Operation of Transformer. It will now be in order to assem- 
ble the separate electrical and magnetic circuits which have been 
explained, to form an idealized transformer, Fig. 9. In this figure 



161 



18 



DESIGN OF SMALL TRANSFORMERS 



NS denotes a round bundle of soft-iron wires forming the core of 
a coil of three turns which has a current sent around it in a 
clockwise direction from a battery of cells denoted by E. This 
will cause the upper end of the core to be an S pole; the mag- 
netic lines will have a downward direction as indicated. Suppose 
a circuit denoted by ABCD is arranged as shown so that a certain 




n 



Fig. 11. Lines of Force Produced by Primary Coil of Transformer 

number of magnetic lines thread downward through it. So long 
as the parts of the arrangement are stationary and the battery 
current does not change in amount, there will he no current in the 
circuit ABCD. Should the battery circuit be opened, as indicated 
in Fig. 10, and the magnetizing current in the coil surrounding the 
iron core becomes zero, the number of lines threading downward 
through the circuit ABCD will decrease rapidly and a current will 



162 



DESIGN OF SMALL TRANSFORMERS 



19 



be induced in circuit ABCD in a clockwise direction according to 
the rules explained. The induced current in ABCD is in a direc- 
tion exactly opposite to the current in the coil surrounding the 
core that produced the magnetic lines that threaded ABCD. The 
coil P surrounding the core NS may be designated as the pri- 
mary, and the circuit ABCD as the secondary, of a transformer in 
which the two electrical circuits are connected together, or inter- 
linked, by the magnetic circuit consisting of the lines of force 
threading through the iron core, the primary, and the secondary. 
If the battery circuit is closed by joining t and f, the current in 




Fig. 12. Principal Parts of Closed-Core Transformer 

P will increase in value, thus increasing the number of lines 
threading through the secondary ABCD, and a counterclockwise 
current will be produced in the closed secondary circuit denoted 
by ABCD. 

Magnetic Leakage. To one at all familiar with the form of 
the magnetic field produced by a coil of wire surrounding an iron 
core, it will be evident that not all the magnetic lines produced 
by the magnetizing force of the ampere turns of the primary will 
pass through the secondary. The actual shape of the magnetic 
field produced by the magnetizing force of the primary coil is 



163 



20 



DESIGN OF SMALL TRANSFORMERS 




Fig. 13. Another Type 
of Closed-Core Trans- 
former. 



indicated in Fig. 11. Those magnetic lines produced by the pri- 
mary coil which do not pass through the secondary are called 
leakage lines. The greater the number of leakage lines, the less 
the useful effect realized. It is advisable, there- 
fore, to so arrange both the primary and the 
^0 ^\ secondary as to have as many of the magnetic 

f I lines as possible thread through both primary 

and secondary. This may be accomplished by 
making the magnetic circuit, or core, a complete, 
or closed, circuit of iron, circular in shape, and 
winding the primary on this in close turns PP, 
Fig. 12. The secondary may be wound over 
the [primary as shown at iS. So far as mag- 
netic efficiency is concerned, the form of trans- 
former illustrated by Fig. 12 is the best that can 
be employed. There are no "joints" to interrupt 
the passage of the magnetic lines, and the path of the magnetic 
circuit is circular. The disadvantage of this form is the increased 
labor required in its building. Each turn of both primary and 
secondary must be threaded through the center hole by hand, 
so, if there are many turns to be wound on the core, the labor 

is considerable. 

Open= and Closed=Core Transform^ 
ers. A transformer having a magnetic 
circuit consisting partly of air and 
partly of iron is termed an open-core 
transformer, Figs. 9 and 10. For a cer- 
tain class of service open-core transform- 
ers are desirable, for example, operating 
certain kinds of wireless apparatus. For 
^^ ^ -^&^^^aB^m ^^^^ classes of power service transf orm- 

nll T^^H ers having magnetic circuits made up 

completely of iron are employed. These 
Fig. 14. Buiit-Up Type of Closed- are Called closcd-corc transformers, 

Core Transformer -p,. -<r»-io ^ -i * 

Figs. 12, 13, and 14. 
Shell and Core Types. Transformers may be so constructed 
that the primary and the secondary windings encircle the iron 
core or so that the iron circuit encircles the windings. When the 




164 



DESIGN OF SMALL TRANSFORMERS 



21 



1 



iron portion is outside, the transformer is a shell type, Fig. 13, and 
when the windings are outside, it is a core type. The core type is 
much easier to construct and is also easier to repair. A partly 
built core transformer is shown in Fig. 14. 

Effects of Number of Turns in Secondary. The amount of 
the induced pressure in a secondary consisting of a single turn 
has been explained and computed. It will be interesting to note 
that if a second circuit, or turn. Fig. 15, be arranged connected in 
series with the first turn, or circuit. Figs. 9, 10, and 11, then the 
total induced pressure of the secondary will be twice what it was 
with the single turn. In 
fact, the secondary termi- 
nal pressure will always 
depend upon the number 
of turns wound to consti- 
tute the secondary coils. 
This fact of secondary pres- ^ 
sure increase holds true for 
any arrangement of turns. 

Effect of Number of 
Turns in Primary. The 
effect of increasing the num- 
ber of turns in the primary pjg jg 
windings is to increase the 
possible magnetizing force due to the primary ampere turns; that 
is> to increase the amount of the magnetic flux with a given 
primary current. 

LOSSES IN TRANSFORMER 
Primary and Secondary Losses. In designing transformers to 
meet any requirement, it is very important to have a very 
thorough knowledge of the different losses and of their relation to 
one another. The losses in a transformer may be classified as the 
'primary loss, the secondary loss, and the core loss. The primary 
loss is expressed in watts and is a true RP loss, depending upon 
the resistance in ohms of the primary windings and the square of 
the current that flows through them, expressed in amperes. The 
power loss in the secondary is expressed in watts and is also an 




Effect of Additional Turns on Secondary 
Terminal Pressure 



165 



22 DESIGN OF SMALL TRANSFORMERS 

RP loss, depending upon the resistance of the secondary windings 
and the square of the secondary current. When there is no load 
on the transformer, that is, when the transformer is giving no use- 
ful output, the current and therefore the power loss in the secon- 
dary is zero; but there may be a very small current and therefore 
a small power loss in the primary. 

Core Loss. Eddy-Current Loss. The loss in the iron core is a 
power loss and is expressed in watts. The core loss may be 
divided into two parts, one called eddy-current loss, the other 
hysteresis loss. A so-called eddy current in any iron core is a 
true electric current induced in the iron by the changing mag- 
netic flux, just as if the iron were the desired conductor. Induced 
currents in the iron core are undesirable, since such currents 
require power input to the device and simply convert the electrical 
input into heat that is not only unavailable for useful electrical 
output but tends to heat the primary and the secondary windings, 
thus increasing the resistance and therefore the power loss in both 
coils. 

The eddy-current loss may be made small by constructing the 
iron core of thin plates of iron that are insulated from each other, 
either by some form of insulating varnish, such as shellac varnish, 
or japan, or simply by the rust that naturally forms over the sur- 
face of iron when exposed to moist air. Iron rust is iron oxide 
and is a nonconductor of electric current. Sometimes varnished 
paper is employed to separate the thin plates; but this is not 
necessary under ordinary conditions and of course increases the 
gross cross-section of the core. 

The eddy-current loss per cubic inch may be expressed by 

in which We denotes the loss in watts; h denotes the thickness of 
the iron sheet forming the core, expressed in mils, or thousandths 
of an inch; /, as before, denotes cycles per second; and /S the flux 
density in gausses, or lines of force per square centimeter of cross- 
section of iron. 

Hysteresis Loss. '^Hysteresis" is a Greek word meaning to 
lag. Hysteresis in an iron core means that the magnetic flux, or 



166 



DESIGN OF SMALL TRANSFORMERS . 23 

lines of force, lag behind the magnetizing force that causes them. 
The nature of the hysteresis effect may be understood by con- 
sidering that whenever a new piece of iron is subjected to a mag- 
netizing force and the magnetizing force is removed, a portion of 
the magnetic flux remains in the iron. In order to remove this 
residual magnetism, another magnetizing force must be applied to 
the iron in a direction opposite to that of the initial magnetizing 
force. In other words, energy has to be supplied to demagnetize 
the iron. In the operation of the transformer the primary current 
is constantly changing not only in value but also in direction, so 
that the magnetizing force of the primary is first in one direction 
and then in the opposite direction. For each single change in 
magnetization from one direction to the opposite one a certain 
amount of power or energy is demanded. The total amount of 
energy thus required per second will evidently depend upon the 
number of reversals that occur each second as well as upon the 
quality of the iron; a soft iron 
retains few lines of force and hence 
requires less applied energy to 
demagnetize it. Fig. 16 will serve 
to represent the simplest form of ^'^- ^^- ^^"5*^° cjci?''""''* ^""""'"^ 
an alternating pressure, current, or 

magnetic flux. The current or the magnetic flux increases 
from zero to a certain positive, or +, maximum, decreases to 
zero, increases in the opposite direction to a negative, or — , 
maximum, and . finally returns to zero, ready to repeat a similar 
cycle of values. One complete cycle of changes in value is repre- 
sented in Fig. 16; if there were 60 such changes in each second, 
the frequency of the current would be rated as 60 cycles. 

Since magnetization affects the molecular constitution of the 
iron of the core, changing the position of the molecules from 
pointing in a certain direction to pointing in the opposite direc- 
tion, it has practically the same effect as the rapid hammering 
of the iron, which would produce heat. 

The loss due to hysteresis may be designated by Wh, and 
its value may be expressed by 



167 




24 DESIGN OF SMALL TRANSFORMERS 

in which Wh denotes loss, in watts, per cubic inch of iron in the 
core; k denotes what is called the hysteretic coefficient, which 
depends upon the kind or quality of iron or steel employed; 
/ denotes the frequency of the applied pressure, primary current, 
and resulting magnetic flux in the core; and /3 denotes the den- 
sity of the magnetic flux, that. is, the number of magnetic lines 
of force that pass through 1 square centimeter of cross-section of 
iron. One hne of force per square centimeter is termed 1 gauss. 
Since the above equation expresses the hysteresis loss in watts 
per cubic inch of iron, the total core loss in any transformer may 
be found by finding the volume of the core in cubic inches and 
multiplying this value by the loss per cubic inch. 

It should be carefully noted that the hysteresis loss depends 
upon the frequency, while the eddy-current loss depends upon the 
square of the frequency. It is evident also that the lower the 
frequency the less will be both the core losses. 

Total Core Loss. The total core loss may be expressed by 
the equation 

POWER EFFICIENCY OF TRANSFORMERS 
Definition of Power Efficiency. The power efficiency of any 
transformer depends upon its design, so that a working knowledge 
of the fundamentals of efficiency is imperative in designing trans- 
formers. The power efficiency of a transformer is expressed as 
a ratio of the useful power output to the total power input; that 
is, if the useful power output is divided by the total power input, 
the fraction so obtained represents the power efficiency. If this 
fraction representing the power efficiency be multiplied by 100, 
the efficiency will be expressed in per cent. The useful output 
of a transformer is obviously the total input minus all the various 
losses, so that the power efficiency rj may be expressed by 

_ Wr-We-Wh-Rph'-RsIs' 

in which Wx denotes the input in watts, We the eddy-current 
loss, Wh the hysteresis loss, Rplp^ the loss in the primary coils, 
and Rsh^ the loss in the secondary coils. 



168 



DESIGN OF SMALL TRANSFORMERS 25 

The power efficiency of a transformer may also be expressed as 

^ EJs 

'^ EsL+RpIp'+RsIs'+Wc 
in which Es denotes the pressure, in volts, between the terminals 
of the secondary; 7s denotes the current in the secondary, meaning 
the useful-load current; Rp denotes the resistance of the primary 
in ohms; Ip the primary current in amperes; Rs the resistance of 
the secondary in ohms; and Wc the core loss in watts. 

The output of a transformer may be expressed as the product 
of the secondary pressure and current whenever the useful output 
is a so-called noninductive load, such as a load of lamps, or a 
simple heater load, such as electric radiators or flat irons. If 
the transformer is employed to furnish power to an a.c. motor, 
the useful output can no longer be expressed by Esis but must 
be expressed by EglsXpower factor (p.f.). The power factor is 
always less than unity for so-called inductive loads, such as 
motors, and for such loads the power efficiency of a transformer 
may be expressed by 

EsIsXp.t 

"^ EsIsXp.t+RpIp'+RsIs'+Wc 

All the losses in a transformer are heat losses; a portion of 
the electrical power input becomes converted into heat that is 
nonavailable for useful output. The losses should therefore be 
kept as small as possible, consistent with cost of construction. 

Variation in Efficiency. The power efficiency of a transformer 
is not constant for different useful outputs, since the losses in the 
primary and the secondary windings depend upon the square of 
the currents in them. The core loss, however, is assumed to be 
constant for all loads, so that it is practically the same with no 
useful output as it is at full-load useful output. This fact has 
a very important bearing on design. 

As just stated, the power efficiency of a transformer is a 
variable and should be at the maximum at normal useful output, 
being less at, say, one-half the normal output and at one and 
one-half times the normal output than at normal. If design is 
based upon power efficiency, this will be true. 

An=Day Efficiency. If a transformer is connected with the 
supply mains during twenty-four hours but is delivering no useful 



169 



26 DESIGN OF SMALL TRANSFORMERS 

output, the energy input is practically equal to the core loss in 

watts multiplied by the time. For example, suppose the core 

loss to be 300 watts; the watt-hour loss during twenty-four hours 

7900 
will be 300X24, or 7200 watt-hours, which equals ^^, or 7.2 

kilowatt hours. While there will be a very small primary cur- 
rent under the assumed conditions, there will be but a small primary 
loss, since the resistance of the primary multiplied by the square 
of a small quantity gives a small quantity as a result. 

The all-day efficiency of a transformer may be expressed as 
the ratio of the useful watt-hour output to the total watt-hour 
input during twenty-four hours. The all-day efficiency is evi- 
dently very different from the power efficiency. If a transformer 
is to deliver a useful output during ^ve hours per day while the 
core loss continues throughout the twenty-four hours, the trans- 
former should be so designed that it will have a comparatively 
small core loss; that is, the copper loss in the primary and the 
secondary windings may be larger than the core loss. 

Ratio of Transformation. The ratio of transformation is the 
ratio of the secondary pressure to the primary pressure. In a 
step-down transformer the ratio of transformation is less than 
unity, while in a step-up transformer it is greater than unity. 
Suppose the applied primary pressure is 110 volts and the second- 
ary pressure is 10 volts; the ratio of transformation is then 

Es _ 10 _1 
Ep 110 11 

Suppose the pressure applied to the primary of a transformer is 
110 volts while the secondary terminal pressure is 22,000 volts; 
the ratio of transformation in this case is 

E. _ 22000 _ 200 _^^^ 
E;~ 110 "X"^^^ 

Regulation of Transformer. By the regulation of a trans- 
former is meant the proportionate change in its secondary termi- 
nal pressure from full load to no load. This is sometimes expressed 
as a ratio as follows: 

Regulation = — ^— — - 
Es 



170 



DESIGN OF SMALL TRANSFORMERS 



27 



in which Vs denotes the no-load terminal pressure of the second- 
ary, which is always greater than the full-load pressure, and Es 
denotes the full-load terminal pressure. These pressures are based 
upon a noninductive secondary load and a constant applied pri- 
mary pressure. 

Summary. Before considering the design of several trans- 
formers to meet specific requirements, it will be well to sum- 
marize the principles already enumerated, which may perhaps be 
best accomplished by referring to Fig. 17, which represents dia- 
gramatically a transformer connected to supply mains having a 
pressure Ep and delivering power to a load of lamps. A watt- 
meter W indicates the true power input under all conditions of 
variable useful output, whether the output is inductive or non- 
inductive. An ammeter Ap indicates the current in the primary; 



S^S) 




Fig. 17. Diagram of Connections for Practical Use of Transformer 

a voltmeter Vp indicates the pressure Ep of the supply mains, 
that is, the primary pressure; and the frequency meter indicates 
the number of cycles per second of the applied pressure. Since 
the diagram represents a step-up transformer, the wide lines 
encircling the core denote the primary windings, while the thin 
lines denote the secondary windings. Voltmeter Vs is arranged 
to indicate the secondary pressure Es; and the ammeter ^s is to 
indicate the secondary current. 

PRACTICAL DESIGN 

DESIGN OF 10,000=VOLT TRANSFORMER 

Specifications. The design of transformers may be considered 
from several standpoints depending upon the nature of the work 
they are expected to perform. The basic fundamental principles 
already considered are applicable under all conditions, however, 



171 



28 DESIGN OF SMALL TRANSFORMERS 

so that a careful study of these fundamentals will enable the 
designer to recognize at once in what detail a change in design 
may be made in order to meet a definite requirement. A brief 
outline applying to the design of a 1-kilowatt, 10,000-volt, step- 
up transformer will first be considered and it may serve as a work- 
ing guide in building other transformers for various pressures and 
outputs. It will be assumed that the applied primary pressure 
is to be 110 volts at 60 cycles. 

Basic Values. If the primary pressure is 110 volts and the 
secondary pressure 10,000 volts, the ratio of transformation is 
therefore 

£p" no ^-^ 

The efficiency of the transformer will be assumed to be 93 per 
cent. Since the output has been assumed to be 1000 watts, the 

input must be , or 1075 watts. Since the output is to be 

u.yo 

1000 watts at 10,000 volts, the secondary current must be 
, 1000 1 



10000 10 



amp. 



Since the input is 1075 watts and the primary pressure 110 volts 
and as the design is not to be strictly rigorous, an assumption of 
10 amperes for the primary current will be close enough 'for the 
purpose. 

Proportioning of Losses. The proportioning of the core loss 
and the copper loss will depend largely upon the service under 
which the transformer is to operate. If, for example, it is to be 
connected with the supply mains during twenty-four hours and 
is to deliver useful power during, say, only three hours, the core 
loss should obviously be much smaller than the power loss of 
the windings. If the transformer is to deliver useful output dur- 
ing all the time it is connected with the supply mains, then the 
core loss will be made relatively much larger than the copper loss. 
For general experimental purposes, the core loss may be made 
equal to the primary copper loss, which proportion will be assumed 
in this design. The next step might be the proportioning of the 
primary and the secondary loss. It will be good practice to have 



172 



Mill 



DESIGN OF SMALL TRANSFORMERS 



29 



the primary loss less than the secondary loss. Let the primary 
loss be assumed to be 20 watts and the secondary loss to be 
35 watts. 

DESIGN OF CORE 

Type of Core. It will be advisable to decide upon the form, 
or shape, of the core before computing its dimensions. A core 
type built up of rectangular plates, Fig. 18, will be assumed. 
The so-called mean length of the iron core is indicated by the 
dotted line. While joints in magnetic circuits are undesirable so 
far as magnetic conductivity of lines of force is concerned, a core 
built up of plates as shown in Fig. 18 so as to ''break joints" 
constitutes a fairly good magnetic circuit and is very easy not 
only to construct but to take apart 
whenever desired. 

Assumed Data for Eddy=Current 
Loss. The equations for the eddy-cur- 
rent loss and the hysteresis loss have 
already been explained, and it will now 
be necessary to assume some of the 
various quantities involved in these 
equations. The equation for the eddy- 
current loss per cubic inch is 



W.= 



16.38 




Fig. 18. BuUt-Up Core of 
Transformer 



The quantities are assumed to be as follows: b equals 



15 



1000 



inch. 



or 15 mils, and b^ equals 225; / equals 60 cycles, and /^ equals 
3600; and M equals 3000 gausses, and M^ equals 9,000,000. As has 
been just stated, the thickness of the iron plates b is assumed to 
be 15 mils; by inspection of the equation, it may be seen that 
as b is made smaller and smaller, the eddy-current loss becomes 
smaller and smaller, all other quantities in the equation remaining 
unchanged. There will be, however, a point beyond which it will 
not be advisable to reduce the thickness of the plates, since for 
very thin plates the insulating varnish covering the plates is too 
thick in proportion to the thickness of the plates. If the plates 
are made very thin, the insulation may be thicker than the plates, 



173 



30 DESIGN OF SMALL TRANSFORMERS 

so that less than one-half of the cross-sectional area of the core, 
Fig. 18, may be iron and more than one-half may be insulation. 
The assumption of 15 mils is a very probable value for b. The 
eddy-current loss in watts per cubic inch may be found by sub- 
stituting the assumed values in the equation as follows: 

We = ^^4§ri^7^777T7^X 225X3600X9 000 000 = 0.01194 watt 

10 000 000 000 000 000 

Ten raised to the sixteenth power is expressed by 1 followed by 
sixteen ciphers. 

Assumed Data for Hysteresis Loss. The hysteresis loss per 
cubic inch is expressed by the following equation: 

in which k equals 0.Q021; / equals 60 cycles; and /^ equals 3000 
gausses. The numerical value of M^-^ may be found by the use 
of logarithms as follows: 

log ;§i-6 = 1.6Xlog /g = 1.6Xlog 3000 

since log 3000 equals 3.477121, therefore 

log ygi« = 1.6X3.477121 =5.5633936 

By hunting up the number having a logarithm of 5.5633936, it 
will be found that 

yg 1-6 = 365000 

Making the proper numerical substitutions in the hysteresis-loss 
equation will give 

= 16.38 X 0.0021 X 6 X 0.365 = 0.07533 watt 

Volume, Length, and Weight of Core. The complete core loss 
per cubic inch will be the sum of the eddy-current loss and the 
hysteresis loss per cubic inch, that is, 0.01194+0.07533, or 0.08727 
watt. From this value the required volume of the core may be 
computed. Since the assumed core loss was 20 watts, it must 
follow that 



174 



DESIGN OF SMALL TRANSFORMERS 31 

It is evident that the mean length / of the core multiplied 

by its cross-sectional area A must be equal to the volume, or 

in symbols 

Axl = 229 cu. in. 

If a numerical value is assigned to A, then the value of I can be 

found. It will be assumed that the cross-sectional area is to be 

square and to be 4.14 square inches, which equals 6.45X4.14, or 

26.70 square centimeters. If the cross-sectional area A is 4.14 

square inches, then 

7 229 .. . 
/=— — = 55 m. 
4 

55 
One-half the mean length will be — , or very nearly 28 inches. 

If the shorter leg of the core can be made, say, two-thirds of the 
longer, the lengths of the longer and shorter strips will be re- 
spectively 17 and 11 inches. 

As iron weighs about 0.27 pound per cubic inch, the core 
will weigh 229X0.27, or 61.83 pounds. 

DESIGN OF PRIMARY 
Number of Turns in Primary. It will be advisable to now 
consider what may be called the fundamental design equation of 
a transformer, which is 

_ \/2"x^XiVpX7r/Xyg 
"^ 10« 

In this equation E^ denotes the pressure applied to the terminals 
of the primary; A denotes the cross-sectional area of the core. 
Fig. 18, expressed in square centimeters; iVp denotes the number 
of turns of wire on the primary coils; tt equals 3.1416; / denotes 
the frequency of the applied pressure; I^ denotes the density in 
gausses, or in number of lines per square centimeter of cross- 
section of core; and v2 equals 1.414. There are three quanti- 
ties represented in this equation that have not been assigned 
numerical values, namely, A, Np, and /^. It is evident that if 
values are assigned to A and M, the required number of turns 
for the primary Np may be found. Furthermore it is evident 
that with any assumed area the greater the value of M the less 
will be the required number of turns. The number of lines 
assumed in designing the core was 3000 per square centimeter, 



175 



32 DESIGN OF SMALL TRANSFORMERS 

and the cross-sectional area of the core was found to be 4.14 square 
inches, or 26.7 square centimeters; using these values in the fun- 
damental transformer equation will give 



110 = 



1.414X26.7X iVpX3.1416x60X3000 



10» 
from which 

110X100 000 000 



N. 



Np = ^ ^^. .^ . ..^. .^, .^ = ^.^ =516 turns 



1.414X26.7X3.1416X60X3000 
which reduces to 

110X1000 ^ 11000 

1.414X2.67X3.1416X6X3 213 
If there are to be two primary coils, one on each limb of the core, 

516 
there will be — ' or 258 turns per coil. 

Length of Primary Wire. The section of the core is nearly 
2X2 inches, the distance around this section must be 4X2, or 
8 inches. If the primary is wound next the core, as is the best 
method, the length of a mean turn must evidently be more than 
8 inches, say 9 inches, and as there are to be 516 turns, the total 
length of the primary will be 516X9, or 4644 inches, which equals 

^, or 387 feet. 

Size of Primary Wire. The primary copper loss Rplp^ was 
assumed to be 20 watts and the primary current 7p 10 amperes. 
Therefore Ip^ equals 100, and the primary resistance Rp may be 
found as follows: 



p 20 2 1 , 
^^ = 100 = I0 = T"^^ 



Since the length of the primary wire is 387 feet, its resistance 

2 
per foot will be -^, or 0.000517 ohm. Consulting Table II, the 

size of wire the resistance of which is nearest to the amount given 

will be found to be No. 8, and therefore a No. 8 B.&S. gage 

wire will constitute the primary. 

Weight of Primary Wire. Since there are nearly 20 feet of 

No. 8 double cotton-covered wire per pound, the required number 

387 
of pounds is — — , or about 19| pounds. 



176 



I 



DESIGN OF SMALL TRANSFORMERS 



33 



o 

c 

bl 

a 

u 

mm O 

— a 

« & 

c 

2» 

a 
& 
< 



1-9 


lis 

1^ 


0004406 

0007007 

0008835 

001114 

001771 

002817 

004964 

007122 

01132 

01801 

02863 

04552 

07239 

1151 

1830 

2910 

4627 

7357 

296 


OOOOOOOOOOOOOOOOOOi^ 


<— 1 




T*H rHOO<N ^ 

oooooooooooooSS^^coo 
ooooooo'ooooooooooorH 


w 


13 


Tt<Ot>.T-i(NiOOOOOOOOOOOOOO 


(MOSTtHi-tOSt^fMC^OCOOi-^Oi-HOOCOt^Or^ 

^S^eo^t-c^OiO^JC2-HOr^GO^-co;-Hoo 
1— 1»— ico'^i>'— ii>.coi>-coo>eoo 

?— ti— lCS|C0»Ot>i— IU3 
1—1 1—1 


o 


Weight per 

1000 Feet of 

D.C.C. Wire 

(lb.) 


(MCOt^(M00r-HCO00 

05iO<MCOO(M<MeOCV|T)HOcOTti 

rHrtHTtH05i-Hb-0iO<MOC000t0C0(Nr-H^OO 


OOOrHO(Nt^^eO(N^OOOOOOOO 
00 »0 ■<*' CO (N T-H 


p^ 


Ids 
5Qo 


CD(NiO(rO(MiOOOi-H-*Oi'*T-iCiCOtOCO'^i-i 

t^Ttl(Mi-lOit^COlO'*CO(N(>5(MT-Hr-I^T-lT-lrH 

;_H^^r-iOOOOOOOOOOOOOOO 


ooooooooooooooooooo 


H 


Area 
(sq.in.) 


O 05 CO lO CO CO 

lOCOrHCOOiTt^COt^COi-ITtl 

COCOrHOOCOCMr^COlOCOiOCO^COCOt^ 

(NOsOi^t^t^COiO^C^t^OOTtiOO Oi^i-i Oi CSit^ 

ooooooooooooooooooo 


Q" 


if 


OOOOOOOOOTtHO-HOO^fN^Ot^OO 


(N 1-1 1-1 rH 


O 





o»OTfio5oooooeooicoi-H05cooaicoooiT-i 


gg^sg^S^j;5^^:2^s^^"^^^ 


PQ 




ouo^oioooooco^^SoSooScooS^ 
2^:=!2goooooooooooo8o 


ooooooooooooooooooo 


< 


B.&S. 
Gage 
No. 


C00005O(M'^C000O<M-rt<COCX)O<N'^C000O 
,_(T-(r-(i-(i-l(M(NC<|(N(MC0C0C0C0C0-^ 



177 



34 DESIGN OF SMALL TRANSFORMERS 

DESIGN OF SECONDARY 
Number of Turns in Secondary. The ratio of transformation 

is , or very nearly 91, and as the number of primary turns 

has been found to be 516 the number of turns on the secondary 
will be 516X91, or 46,956 turns. 

Resistance of Secondary. The loss in the secondary has been 
assumed to be 35 watts and the secondary current will be assumed 

to be — ampere. The resistance of the secondary will therefore be 
E. = -^ = 4235 ohms 



11, 

Length of Secondary Windings. The length of secondary 
wire depends upon the assumed current density in this wire. By 
current density is meant the number of amperes per square inch 
of cross-sectional area of conductor. The current density allow- 
able depends upon the method of winding. For ordinary closely 
wound coils the current density should not exceed 500 amperes 
per square inch of cross-section of copper conductor for continu- 
ous service. If the transformer is used intermittently, the current 
density may be increased to as high as 1000 amperes per square 
inch. Assuming. 500 amperes per square inch, since the secondary 

current is — , or 0.09 ampere, the area of the secondary wire 

will be 



^-i)'^T(ra=54)=«-°°°^«^^-'"- 



1 



Consulting Table II, the size having a cross-sectional area near- 
est to this is No. 26 B.&S. gage, the resistance of which is 
0.04075 ohm per foot. If No. 26 B.&S. gage is used and the total 
resistance of the secondary is 4235 ohms, the length required for the 

4235 
secondary will be tt-^ttt^, or 103,923 feet. This length of second- 

ary should be so wound on forms of the proper size that the total 
number of turns shall be at least 90X516, or 46,440 turns. 

The required length of a mean turn may be found by divid- 
ing the length of the secondary by the required number of turns, 



178 



DESIGN OF SMALL TRANSFORMERS 35 

103923 
which gives , or 2.2 feet, which equals about 26.4 inches. 

Since the length, or circumference, of a mean turn is known, its 

radius may be ascertained as follows: 

27rr = 26.4 

26.4 26.24 . , . 
^ = -;; — = ^ ^^ =4.1 m. 

27r 6.28 

Weight of Secondary Windings. Before proceeding with the 
considerations pertaining to the methods of winding the coils 
and assembling the core, it will be in order to compute the 
weight of the secondary wire. This may be done if the number 
of circular mils in the cross-sectional area of the secondary wire, 
the length of the secondary, and the weight of a circular-mil foot 
are known. The weight of a round copper wire 1 foot long and 
having a cross-sectional area of 1 circular mil, that is, 1 circular- 
mil foot, is 0.000003027 pound. Since B.&S. gage wire No. 26 
is 254.1 circular mils in area and the length of the secondary is 
103,923 feet, the weight of the secondary is 0.000003027X254.1 X 
103 923, or 79.9 pounds. As this is figured for bare wire, a com- 
parison should be made with the weight of double cotton-covered 
wire based on the values given in Table 11. By consulting this 
table, it will be seen that the weight of No. 26 double cotton- 
covered wire is 0.822 pound per 1000 feet. This would make 
the weight of 103,923 feet equal 103.923X0.822, or 85.4 pounds. 

CONSTRUCTION OF TRANSFORMER 
Constructing Core. A few general directions which apply 
to the building, or assembling, of the iron core are now in order. 
The iron strips may be obtained cut to the proper size or they 
may be cut by hand from ordinary stovepipe iron, the so-called 
Russia iron. Only two sizes of these sheets, Fig. 18, are needed, 
and they may be assembled as indicated in Fig. 19. The strips 
should be carefully inspected and all rough edges smoothed with 
a/ fine file or sandpaper, after which both sides and the edges of 
each strip should be carefully coated with shellac or an equally 
good varnish, employing a brush. While the varnished strips 
are drying, which will require at least twenty-four hours in a dry 
temperature of 70° F. the wooden base and the necessary blocks 
and strips may be prepared as shown in Figs. 19 and 20. 



179 



36 



DESIGN OF SMALL TRANSFORMERS 



After the core is built up to the proper thickness, a wooden 
strip AB, Fig. 20, may be so screwed to the wooden base CD as 
to hold the core firmly in position. If the core is large and heavy, 




Fig. 19. Method of Assembling Transformer Core 

bolts should be used instead of screws to hold AB and the core. 
By holding the core as indicated, the primary and the second- 
ary coils may be slipped on over the two legs of the core 
and the end strips inserted to complete the iron core. This 
method of building and arranging the core makes assembling 
very easy and facilitates the making of any desired change in 




Fig. 20. Method of Clamping Transformer Core during Construction 

the coils at any time. The height of the two supporting blocks 
will need to be ascertained from the size of the transformer to 
be built. The blocks should hold the core above the base at a 



180 



DESIGN OF SMALL TRANSFORMERS 



37 



sufficient height to allow the primary and the secondary coils to 
be slipped onto the core. Soft pine is the best material to use for 
the wooden base and blocks, as this wood retains its shape much 
better than does any other kind under the ordinary climatic con- 
ditions of temperature and moisture. 

The number of iron strips necessary for the building up of 
the core will obviously depend upon the thickness of the material. 
15 



If the metal is 



1000 



inch thick and the core is to be 2 inches thick, 



or 



2 

15 
1000 
required. 



2000 
15 ' 



or 133 strips placed one above another will be 



This means that there must be 2X133, or 266 longer 




Fig. 21. Winding Primary Coil 



strips, and the same number of shorter strips. If the thickness 



of the iron used is greater than 



15 



1000 



inch, the required number 



of strips will be less. It will not be advisable to use iron that is 
15 



thinner than 



1000 



inch. 



Constructing Primary, The primary may be wound on round 
tubes made of two layers of cardboard glued together. Fig. 21. 
The tubes may be made up over a cyjindrical form such as is 
represented in Fig. 22. This form is made with a diagonal cut 
as indicated at cc^ and may be so constructed as to serve for 
winding the primary, if provided with a handle h. This cylinder 
should be made at least 2 inches longer than the tube on which 
the primary is wound in order that it may be unscrewed at the 



181 



38 



DESIGN OF SMALL TRANSFORMERS 




fJA 



Fig. 22. Method of Making Cardboard 
Tubes 



end marked 8 and at the opposite end to allow the two wedge- 
shaped portions to be pulled from the finished coil. This same 
cylinder might be adapted for winding the secondary coils if 

they are to be wound by hand, 
but the easiest method of wind- 
ing the secondary is in a lathe. 
Before winding the first 
layer of wire on the cardboard 
tube, at least three strips of 
sticky tape should be laid on the 
tube so as to extend over the ends 
of the tube for about 3 inches, 
Fig. 2L After the first layer of wire is wound on the tube, the 
free ends of the strips of tape should be brought over the top of 
the first layer, and other strips of tape laid on so as to bind in 
the second layer of the primary when wound on. The card- 
board tube cannot be over 16 inches in length for the trans- 
former under discussion. It will be advisable not to wind the 
primary turns of wire nearer than 1 inch from each end of the 
tube, which would make the coil 14 inches long. 

Diameter of Tube. The wind- 
ing of the primary on a round 
tube to be slipped over an 
iron core having a square cross- 
section implies that the diam- 
eter of the tube must be some- 
what greater than the diagonal 
of the square section of the core. 
In Fig. 23 the core is 2 inches 
on a side and the diagonal DB 
is found as follows: 




Fig. 23. Method of Finding Diameter of 
Tube to Encircle Square Core 



DB=VAB'-]-AD' 
Since AB equals AD 



therefore 



DB=V'2AB''=-\/2x2x2=\^8 = 2.82 in. 



Therefore the diameter of the winding cylinder shown in Fig. 22 
should be about 2.9 or even 3 inches in diameter. The tube form 



182 



DESIGN OF SMALL TRANSFORMERS 



39 



allows air to circulate freely about the iron core, thus tending to 
keep it cool. If desired, a tube having a square section may be 
made to hold the primary windings and to slip over the legs of 
the iron core. Figs. 24 and 25 illustrate the method of building 
a winding device and coil forms for a primary such as has been 
described. 

Thickness of Layers of Primary. The wooden ends of such 
a form should not be greater in diameter than is necessary to 
hold the required number of layers of primary. The number of 
layers may be calculated from the number of turns as follows: 




Fig. 24. "Construction of Form for Primary Windings 

The total number of primary turns was found to be 516, which 

516 
will require -— , or 258 turns for each primary coil. The size 

of wire used in the primary coil is double cotton-covered No. 8 
B.&S. gage, which has a diameter of 0.142 inch, column F, 
Table II. If it is decided to make the length of the primary coil 
(not the length of the wire) 14 inches, the number of turns of 



No. 8 per layer will be 



14 



0.142 



or 98 turns. If there are 98 turns 



per layer, there will be 3X98, or 294 turns, in three layers. In 
practice this number could not be wound on, due to slight irregu- 
larities. However, as 294 is considerably in excess of the required 



183 



40 



DESIGN OF SMALL TRANSFORMERS 



number of 258 turns, three layers will accommodate the proper 
number. The thickness of three layers will be 3X0.142, or 
0.426 inch, so that the width of the shoulder of the wooden ends 
for the primary coil forms should be 0.42 inch. The ends may 
be made of wood i inch thick and may be glued to the card- 
board cylinder. 

Winding Secondary. The secondary will need to be wound 
in permanent forms or in forms that may be readily taken apart 
to allow the removal of the wound coil. Since the pressure of 
the secondary is very high, 10,000 volts, the secondary must be 




Fig. 25. Device for Winding Primary Coil 

wound in a number of sections, or coils, so that the pressure 
per coil is greatly reduced. 

Since the secondary will encircle the primary, it will be dis- 
tributed over the two legs of the transformer as is the primary. 
The pressure of 10,000 volts will therefore be divided into 5000 
volts per leg, and if ten coils per leg are adopted, then the per- 

coil pressure will be — -, or 500 volts. The total number of 

secondary turns being 46,440, the number of turns per leg will 
be 23,220 and that per coil will be 2322. The length of the 
winding space is, say, 14 inches, and if the thickness of the 
sides of the coil forms is | inch, the space occupied by the sides 



184 



DESIGN OF SMALL TRANSFORMERS 



41 



of the ten coil forms will be 20 Xi, or 2.5 inches, leaving 14—2.5, 
or 11.5 inches for winding distance. To find if there is sufficient 
winding space inside the core legs to accommodate the required 
number of secondary turns, it will be necessary to refer to col- 
umn F, Table II to ascertain the diameter of No. 26 double- 
covered wire, which is 0.0249 inch. The number of possible 



turns in one layer of secondary will be 



11.5 



, or 461 turns. If 



0.0249' 

there are 461 turns in one layer and 23,220 turns are demanded, 

23220 
the number of layers must be — — -, or 50 layers, and 50x0.0249 

461 




Fig. 26. Dimensions of Parts of Transformer 

equals 1.245 inches. As there is a distance of 9 inches between 
the two legs of the core, there are 4.5 inches for the turns of 
both primary and secondary on each leg. The outside diameter 
of the completed primary should not be over 6 inches, which 
allows 2 1 inches for the secondary windings of each leg. Fig. 26. 

Paraffin-Dipped Coils. Having ascertained that there is ample 
space in which the wires of the primary and the secondary may 
be wound, the methods of winding the secondary and the con- 
struction of forms to hold the secondary may be considered. 

If the separate secondary sections are not to be wound in 
permanent forms, they must be wound in a form so constructed 
as to allow the windings of the coil to be bound together, by 
string or tape, until they can be dipped in melted paraffin 



185 



42 



DESIGN OF SMALL TRANSFORMERS 




Fig. 27. Winding Device for 
Winding Second<iry Coils 



for a time, then removed and cooled, which process holds the 
turns of wire firmly in place. A winding device for this method 
is illustrated in Fig. 27. Four narrow pieces of tape may be 
passed through the four slots and held in 
position by tacking their ends to the face 
of the winding form before beginning to wind 
the coil. After the coil is wound, the 
tacks may be removed and the free ends of 
the tape carried through the upper end of the 
slots and firmly tied together, binding the 
turns of wire so that the outer disc may be 
removed and the coil taken from the form. 
A revolution indicator may be attached. 
Fig. 27, to register the number of turns that 
are wound on the coil. A cross-section of the winding device is 
shown in Fig. 28. In winding, it is not advisable to wind the 
fine wire too tightly. A fairly loose winding is preferred. 

The best method for impregnating the wound coils with paraf- 
fin is to provide a round metal dish about 1 inch larger in diame- 
ter than the diameter of the 
finished coil and two or three 
times as deep as the coil is 
thick. Have such a dish 
about one-half full of very hot 
melted paraffin, almost boil- 
ing, and gently insert one of 
the coils. It may be well to 
make a wooden rack of small 
|-inch sticks to prevent the 
coil from coming into contact 
with the bottom of the metal 
dish. The best method of 
heating hot paraffin is to place 
the dish on the top of an 
ordinary cook stove in which 
a coal fire is burning. The dish may be placed on a gasoline 
torch, but a cast-iron stove cover or thick iron disc should be 
placed between the flame and the dish. Care should be exer- 




Fig. 28. Cross-Section of Winding Device 



186 



I 



I 



DESIGN OF SMALL TRANSFORMERS 43 




cised not to scorch the insulation of the wire in impregnating 
the wire with paraffin. A collection of finished secondary coils 
impregnated with paraffin is shown at e and / in Fig. 29. 

Coils on Permanent Forms. If the secondary coils are to be 
wound on permanent forms, such forms may be made of thin soft 
pine or of thick cardboard, Fig. 30. The hub H may be turned in 
a lathe and the discs D cut out from wood or cardboard not 
thinner than | inch and attached to the hub with ordinary glue. 
The dimensions of such forms may be readily figured from data 
already given. 

A small hole, as at A, Fig. 30, is made through the side of 
such a form just above the hub so that one end of the secondary 
wire may be inserted from the inside, allowing it to protrude from 
the hole for about 6 inches to provide a sufficient length for con- 
necting with the adjacent coil when assembled. Another small hole, 
as at By should be made near the outer 
edge [of the disc to allow the other end 
of the finished coil to protrude 6 inches. 

Method of Connecting Coils. One 
half of the secondary coils should be 
wound in one direction and the other 

half in the other direction so that two Fig. 29. Secondary Coils after 

Being Treated with Paraffin 

coils may have their two inside ends con- 
nected together and form a continuous winding in the same direc- 
tion from the free end of one coil to the free end of the other. 
This is an important consideration and applies to the paraffin- 
impregnated coils. The winding of each coil may be in the same 
direction so far as the turning of the winding device is concerned, 
but the inner end of every other coil that is wound should project 
from the opposite side of the form, or coil. For example, suppose 
one coil is wound in a clockwise direction as one faces it, with its 
inner ends protruding toward the observer. The next coil should 
be started with its inner end protruding from the side of the form 
away from the observer and, as before, should be wound clock- 
wise. If the two coils are placed face to face with their two inner 
ends fastened together, they will constitute a continuous winding 
in the same direction. Likewise if two such coils have their outer 
ends connected together, they will constitute a continuous winding 



187 



44 



DESIGN OF SMALL TRANSFORMERS 



in the same direction. The reason for this method of connection 
is to prevent an excessive pressure between any adjacent layers or 
turns; the terminals of each coil are as far from each other as pos- 
sible by this arrangement. 

Insulating Coils. The question of insulation is a very impor- 
tant one in the design and the operation of transformers. The 
insulation between the primary and the secondary needs particular 
attention in high-pressure transformers. If the secondary pressure 
is above 10,000 volts, a tube of good insulating material, such as 




SECT/O/V £-E 




Fig. 30. Method of Winding Secondary Coils 



so-called empire cloth, having walls \ inch thick, should separate 
the primary from the secondary. Such a tube is shown at t and 
i' , Fig. 31. The same figure also shows a primary having several 
terminals for connecting with service mains having different line 
pressures, and at e and e* may be seen a number of the paraffin- 
impregnated coils assembled. When these coils are used, they 
must be carefully separated from each other by insulating washers 
of some good insulating material. Washers and also tubes may be 
built up by using a number of layers of paper carefully shellacked 
and pressed together before the shellac becomes dry. Fig. 32 
shows a completed high-pressure transformer for 10,000 volts 



188 



DESIGN OF SMALL TRANSFORMERS 



45 



secondary pressure. A number of primary terminals that allow 
the transformer to be connected across service mains of various 
pressures are shown in this illustration also. 




Fig, 31. Transformer Coils and Insulators 

Use of Oil as Insulator. So-called transformer oil, which is an 
oil obtained from petroleum, is an excellent form of insulator, as it 
not only insulates but also carries away a considerable quantity of 
heat by convection currents, or circulation, from the core and coils 
to the enclosing casing and the outside air. If a spark discharge 
takes place in an oil-insulated transformer owing to a sudden and 
temporary increase in pressure, the oil, being a fluid insulator, 
immediately flows in and mends the broken-down insulator, thus 
acting as an automatic mender. Oil insulation is to be recom- 
mended for high-pressure transformers, say 5000 volts and above. 
The use of oil implies a tight 
box, or container, to prevent 
oil leakage. The box should 
be somewhat larger than the 
transformer and may be made 
of wood, if desired, with its 
joints made oil tight by glu- 
ing. Such a box should re- 
ceive several coats of shellac 
varnish both inside and out 
and should be thoroughly 
dried before being used. Another style of container may be made of 
wood carefully nailed together, covered over outside by ordinary roof- 
ing tin, and with carefully soldered joints to make it oil tight. This 




ZtA. 



Fig. 32. Completed High-Pressure Transformer 



189 



46 



DESIGN OF SMALL TRANSFORMERS 



makes a comparatively cheap and very durable and effective 
transformer case. A faucet may be soldered to the container near 
the base to facilitate the removal of the oil when desired. Such 
a container is illustrated and more fully described on p. 87. 

MISCELLANEOUS DETAILS 
Connecting Coils for Different Pressures. If two transform- 
ers like the one just described be connected as shown in Fig. 33, 
having their primaries Ep connected in parallel across the service 
mains and their secondaries Es in series, the secondary terminal 
pressure of the arrangement will be 20,000 volts. When connect- 
ing up for this result, care must be taken that the proper second- 











f § 






Ss'^^O.OOO VOlTsS 



Fig. 33. Transformers Connected in Series 

ary terminals are connected together. The two secondaries have 
relative positive and negative terminal polarity, so that if incor- 
rectly connected together, the net secondary pressure might be 
zero. This may be tested by inserting a very small fuse in the 
secondary circuit so as to short-circuit through the fuse. A short 
piece of No. 40 B.&S. iron wire will act as a fuse, if fine fuse 
wire is not available. If the secondaries are properly connected, 
the fuse will blow. If incorrectly connected, the fuse will not 
blow. A fuse may be employed in the primary circuit for the 
same test if desired. The primaries should be protected by fuses 
of the proper size at all times. A too heavy fuse in the primary 
circuit might not answer for the test just outlined. 



190 



r 



DESIGN OF SMALL TRANSFORMERS 47 



It is evident that three transformers might have their pri- 
maries in parallel and their secondaries in series to give a second- 
ary pressure of 30,000 volts, and four to give 40,000 volts. 

The two primary coils should obviously be properly connected 
so as not to oppose each other in their action. It may be noted 
that if (Jesired, each primary coil might be connected with a 
55-volt 60-cycle supply main and cause the secondary to give 
10,000 volts. If the two primary coils properly connected together 
in series are supplied with 55 volts 60 cycles, the secondary pres- 
sure will be about 5000 volts. 

Step=Down Transformers. Relation to Step-Up Transformers, 
The step-up transformer described may be used as a step-down 
transformer, provided a pressure of 10,000 volts be applied to the 
terminals of the fine-wire coils, or to the secondary of the trans- 
former designed. If a pressure of 10,000 volts is applied to the 
secondary, or to the fine-wire coils, then the fine-wire coils become 
the primary and the coarse-wire coils the secondary. If, therefore, 
a 60-cycle alternating pressure of 10,000 volts be applied to the 
fine-wire coils of this transformer, 110 volts 60 cycles may be 
obtained from the terminals of the coarse-wire coils. Every step- 
up transformer may therefore become a step-down transformer if 
the proper pressure is applied to it. 

Pressure Less than 110 Volts. As it is often desired to obtain 
pressure very much less than 110 volts, some consideration will be 
given to the design of transformers for obtaining pressure less than 
110 volts from 110- volt supply mains. The core and the primary 
of the transformer already computed may be used for the basis of 
a step-down transformer giving an output of 1 kilowatt and a 
pressure and a current of any proper corresponding values. Sup- 
pose it is desired to obtain a pressure of 11 volts, this at 1 kilo- 
watt output will mean — pj— j ^^ ^^-^ amperes, which may be 

considered as 91 amperes. The primary contains 516 turns, so the 

516 
secondary, as a step-down to 11 volts, will need only — -, or about 

47 turns. 

The secondary wire must be large enough to carry 91 amperes 
without overheating. Assuming a value of 1000 amperes per 



191 



48 DESIGN OF SMALL TRANSFORMERS 

square inch, the required cross-sectional area of conductor will be 

-, or 0.091 square inch. Consulting column E, Table II, it will 

be seen that seven No. 8 wires will have a combined area very 
nearly equal to 0.091 square inch, to be exact, 0.012969X7, or 
0.090783 square inch. Therefore seven No. 8 wires may be used. 
From column B, Table II, it is seen that the diameter of bare 
No. 8 wire is 0.1285 inch, therefore if seven No. 8 wires lie side 
by side, one turn will require 7X0.1285, or 0.899 inch of space. If 
24 turns are put on each leg of the transformer, making 48 turns 
instead of 47, then a space of 0.899X24, or 21.58 inches, will be 
required. The seven wires should be carefully soldered together at 
the ends, and if desired, they may have a large binding post 
soldered to them. 

Radiation from Windings. The winding of the seven wires to 
form a secondary brings out a very important principle of heat 
radiation in transformer design. Heat radiates from the surface of 
a conductor, and the amount of radiation depends upon the area 
of its surface. Conductivity, however, depends upon its volume. 
A round wire has an area of surface depending upon its circum- 
ference, which depends upon its diameter, while its volume depends 
upon its cross-sectional area, which depends upon the square of its 
diameter. Therefore a number of small wires which have, col- 
lectively, the same conductivity as one large wire have a larger 
radiating surface. 

This may be explained by referring to Fig. 34. Suppose a 
square A BCD to have a large circle with a radius R inscribed in 
it and to have also four equal small circles (each with a radius r 

which is equal to — ) tangent to each other and to the sides of 

the square as shown. The circumference of one small circle 
equals one-half that of the large circle, and the sum of the cir- 
cumferences of the four small circles is twice the. circumference of 
the large circle. The area of one small circle is equal' to one- 
fourth that of the large circle, and therefore the combined areas of 
the four small circles equal the area of the large circle; that is, the 
four small circles might represent four small conductors whose 
combined radiating surface is twice that of the large circle, or 



192 



DESIGN OF SMALL TRANSFORMERS 



49 



conductor, and whose combined conductivity is the same as that 
of the large conductor. This may be proved by mathematics as 
follows : 

I Circumference of large circle = 2 ttR 
Circumference of 1 small circle = 2 7rr 
Sum of circumferences of 4 small circles = 8 ttt 
Since R equals 2r 

Circumference of large circle = 27rX2r = 47rr 
Therefore 

Sum of circumferences of 4 small circles = 2 circumferences of large circle 
Area of large circle = ttR^ 
Area of 1 small circle = ttt^ 
Sum of areas of 4 small circles = 4irr^ 
Since R equals 2r and R^ equals 4r2 

Area of large circle = ir^r^ = 4 ttt^ 
Therefore 

Sum of areas of 4 small circles =area of large circle 

If these circles denoted round wires, it is evident that while 
the conducting cross-section of the four small wires is the same as 




Fig. S4. Relative Areas of Circles 



that of the large wire, the radiating surface of the four small 
wires is twice that of the large one. If nine equal small round 
wires have the same conductivity as a single large round one, they 



193 



50 DESIGN OF SMALL TRANSFORMERS 

will have three times the radiating surface, since the diameter of 
each of the nine small wires is one-third that of the large wire. 

Design for Very Large Current Transformers built for large 
currents should have their conductors made up of several small 
wires, or in the form of a cable, in order to better radiate the heat 
as well as to make a more flexible wire to manipulate in winding 
the coils. It may be that for welding operations a very large cur- 
rent may be desired, and for an extreme condition the maximum 
current possible from the 1 -kilowatt transformer will be computed. 
It will be supposed that the primary as designed is to be connected 
to a 100-volt 60-cycle supply main, with the secondary to have 
only one turn. Since there are 516 primary turns, the per-turn 

primary pressure is — -, or 0.21 volt. The pressure of the one 

turn of secondary will be about 0.2 volt, and if the output is 1000 

watts, the current would be — ^, or — - — , or 5000 amperes. 

To" ^ 

The sectional area of the single turn would have to be large 
enough to carry 5000 amperes. If a single wire is used for the one 
turn, the current density in it cannot be 1000 amperes per square 
inch because of the small radiating surface compared with its con- 
ducting area. The current density in this case should be about - 

• 1. • • 5000 ,^ . , 

500 amperes per square mch, requirmg ^7—, or 10 square mches. 

500 

The single turn could therefore be 10 inches wide and 1 inch 
thick. This single turn could be made of 16 strips of copper, 
each 10 inches wide and re inch thick, the ends being riveted 
together with a number. of copper rivets and then riveted to suit- 
able clamps for clamping the iron to be welded. 

The method of procedure has beea sufficiently outlined so 
that any desirable departure from the fixed conditions mentioned 
may be made by anyone to obtain pressures and currents within 
the range of values computed. 

Importance of Frequency in Transformer Design. As many 
questions are constantly arising regarding the changes in design 
necessary in order to construct a given transformer for circuits 
having different frequencies, a few considerations relative to this 
matter will be in order. As the primary of any transformer acts 



194 



DESIGN OF SMALL TRANSFORMERS 51 

as a simple "so-called "choke" coil, the current (primary) in which 
may be expressed by 

VR'^{2TfLy 

it is evident that if the applied pressure E is kept constant, the 
current I will vary inversely as the frequency. This may be better 
illustrated by squaring both members of the last equation and 
arranging the factors as 

E' 



P = 



m+4.7r''pD 



For any given primary the resistance R is fixed when the coil is 
wound, and if the flux density in the core is kept constant, the 
inductance L will remain very nearly constant. This relation will 
be best illustrated by a numerical example. 

Suppose the applied pressure is 110 volts at 60 cycles, the 
primary resistance is 1 ohm, and the coefficient of inductance is 25 
henrys. The current will then be 

110 



Vl +4x2x602x252 

110 

Vl +39.4784X3600X625 

110 
^9424 



0.011 ampere 



Suppose the frequency is changed to 120 cycles per second, all 

other factors remaining as before. The current will now be 

expressed by 

110 



Vl+47r2xl202X252 

no 

Vl +39.4784X14400X625 

110 

^8849 



= 0.0058 ampere 



or slightly over one-half what it was when the frequency was 60 
cycles. 



195 



52 DESIGN OF SMALL TRANSFORMERS 

It is evident that a transformer designed for operation on a 
120-cyele circuit will receive about twice as much current when 
used on a 60-cycle circuit if the numerical value of the applied 
pressure is the same in both cases. This at once implies that 
there should be twice as many turns on the transformer designed 
for 120 cycles as on the one designed for 60 cycles if the applied 
pressure is 110 volts in each case. This again shows how little 
bearing the ohmic resistance of the transformer has relative to the 
frequency. A very close approximation to the current in a trans- 
former may be expressed by 

E E 



1 = 



V{2TfLy 27rfL 



If L remains constant, then the ratio of the currents in a 
given transformer when connected with 110 volts 60 cycles and 
with 110 volts 25 cycles may be expressed by 

E 




r 27r60L 25 _ 5 __^,,^ 

r~ E "60 12 "''•'''' 

27r25L 

r=^roir=^- r=2Ar 

The current on a 25-cycle circuit will be 2.4 times as much as on a 
60-cycle circuit. 

Two transformers each designed for a 110- volt 120-cycle cir- 
cuit may have their primaries connected together in series and 
their free primary terminals connected with a 110-volt 60-cycle 
circuit and thus have their normal rated current in their primaries. 
Their secondaries may be connected in parallel and thus give the 
same secondary pressure as would each transformer when connected 
on a 120-cycle circuit. The combined power output of the two 

! transformers will be twice that of each one separately regardless 

of the way they are connected. 

Transformer for Use 07i Circuits with Different Frequencies. 
The design of a step-down transformer having a primary for 110 

I volts and with taps so arranged as to allow it to be connected to 

circuits having different frequencies will be outlined. As the dif- 



196 



^Sent 



DESIGN OF SMALL TRANSFORMERS 53 



•ent frequencies in commercial operation are limited to 25, 40, 
60, and 125, only these will be considered. 

The core will consist of iron strips in two sizes, the longer 
ones 6J inches long and 1 inch wide and the shorter ones SJ 
inches long and 1 inch wide. Enough of these should be provided 
to build up a core 1 inch thick, which would mean that the cross- 
section of the core is 1 inch square. The number of strips required 
will depend upon their thickness. These should be arranged as 
indicated in Figs. 18 and 19. The primary is of No. 25 B.&S. 
gage, double cotton-covered copper magnet wire, arranged on 
forms as shown in Fig. 24 and wound as follows: 5 layers per 
form, 130 turns per layer. Bring the ends of each layer out 
through small holes in the ends of the form. 

When all layei:s of each form are properly connected together 
in series and both coils thus formed connected together in series, 
the primary may be connected across 110-volt mains at 25 cycles. 
If only two layers of each form are connected together in series 
and the two coils thus formed connected together in series, the 
arrangement may be connected with 110-volt mains at 60 cycles. 
If one layer on one form is connected in series with one layer on 
the other form, the two may constitute a primary to be connected 
across 110-volt mains at 125 cycles. If four layers of each form 
are connected together in series and are in series with four series 
connections of the other form, the arrangement may be employed 
on 110-volt 40-cycle mains. By connecting three layers of one 
form in series and in series with two series layers of the other 
form, the arrangement may be adapted for 110 volts and 40 
cycles. For a circuit of 110 volts 100 cycles, two layers of one 
coil may be connected with one layer of the other coil, all con- 
nected in series. 

DESIGN OF 2200=VOLT TRANSFORMER 

Specifications. As pressures in the vicinity of 2200 volts are 
often desirable, the method of designing a transformer that will 
give a useful output of 2 kilowatts at 2200 volts secondary pres- 
sure, with an applied pressure of 110 volts at 60 cycles, will now 
be considered. 



197 



54 DESIGN OF SMALL TRANSFORMERS 

Number of Turns in Primary. Employing the fundamental 
equation for the number of turns in the primary coils, after 
assuming a flux density of 3000 gausses, a cross-sectional area for 
the iron core of 88.8 square centimeters (or 13.76 square inches), 
and 60 cycles for the frequency, the following equation results: 



iVp = 



lO'XEp WXllO 



V^TfAM \/2X3.1416x60x88.8X3000 
11000 000 000 



7100482.8416 



154.5, or 154 turns 



In which Np denotes the number of turns in the primary; Ep 

denotes the pressure, in volts, applied to the primary terminals; A 

denotes the cross-sectional area, in square centimeters, of the iron 

core; / denotes the frequency of the applied pressure in cycles per 

second; and /B denotes the density of the magnetic flux in the iron 

core in gausses, that is, in number of magnetic lines of force per 

square centimeter. In order to have the same number of turns on 

each of the two coils in which the primary will be wound, the 

154 
number of primary turns will be called 154. This will allow — , 

or 77 turns for each of the two primary coils. 

Efficiency of Transformer. In the designing of all kinds of 
power devices, the most important consideration is the efficiency 
of the device as a powder machine. While high power efficiency is 
not ahcays the first consideration in the design of machinery, the 
design of transformers is nearly always based on this consideration. 
While a machine is designed to perform a certain function, it is 
always to be desired that the machine perform its function as 
efficiently as possible from the standpoint of power input and output. 

While the experimenter who constructs a transformer for 
pleasure does so from other considerations than simply that of 
high efficiency, yet when calculations are to be made that will 
determine the size of the various parts of a proposed transformer, 
a very safe method of procedure is one based on power efficiency. 
If power efficiency is to be the basis of design, a thorough knowl- 
edge of its fundamental principles is essentiaL • 

One fundamental fact is that the larger the transformer, the 
greater will be its power efficiency. This means that a trans- 



198 



DESIGN OF SMALL TRANSFORMERS 55 

former for an output of 10,000 kilowatts will have a higher 
efficiency than will a transformer operated from the same circuit, 
built exactly the same with the same grade of material, whose 
output is, say, only 1 kilowatt. 

It is evident that some knowledge of the efficiency of trans- 
formers must have first been obtained as the result of experiments 
which gave information relative to the kind of material used in 
the iron core, the kind of material used in the copper circuits 
constituting the primary and the secondary, the proper form to 
give to the various parts, and their proper relations to one another. 

Much valuable information may be obtained by consulting the 
results of tests performed with transformers by reliable experi- 
menters, and the equations that will be employed in considering 
the design of the 110- volt to 2200-volt transformer are those that 
have been found to give the best results in ordinary practice. 

When about to decide upon the eflficiency of the proposed 
transformer, it will be necessary to know if a 2-kilowatt trans- 
former would be considered a large or a small transformer. Such a 
transformer is really a small transformer compared with many that 
have been constructed, and an assumption of 80 per cent efficiency 
will be a proper one. 

8 
Power Input. If the efficiency is 80 per cent, only — of the 

power input to the transformer will be available for useful power 

2 
output. In other words, — of the input goes into power loss. 

Accordingly, if the useful power output is to be 2 kilowatts, or 
2000 watts, the ratio of the power input to the useful power out- 
put is as follows : 

10 

^ . , 10 10^10 10 

Power mput = — =-X-=- 



2000 ^ 



From which is obtained 



Power input = 2000X^ = 250X10 = 2500 watts 



Primary Current. Relation between Alternating Current and 
Pressure. According to all the laws governing the power in d.c. 



199 



56 



DESIGN OF SMALL TRANSFORMERS 




Fig. 35. 



Variation of Pressure during 
One Cycle 



circuits, the product of the pressure times the current in any circuit 
is always equal to the power, in watts, in the circuit. This, however, 
does not hold for a.c. circuits; so it cannot be said that the pressure 

that is applied to the terminals 
of the primary of a transformer, 
multiplied by the current in the 
primary coil, will give the input 
to the transformer in watts. 

Since the pressure that is 
applied to the terminals of the 
primary is an alternating pres- 
sure, if its various instantaneous 
values, in volts, could be measured and then properly plotted to 
scale, the curve would have in general the shape or form shown in 
Fig. 35. Under ordinary conditions the shape of an actual pressure 
curve will never be as regular as illustrated, but every alternat- 
ing pressure passes through a cycle of values, both positive and 
negative, in a certain definite interval of time. A 60-cycle alter- 
nating pressure is one that completes 60 of the cycles illustrated 
in Fig. 35 in one second. The alternating current in a circuit 
produced by an alternating pressure may be represented by 
Fig. 36. While an ordinary alternating current will not have a 
form quite as regular as depicted, it will in general conform to the 
shape shown. The frequency of the current will be exactly the 
same as the frequency of the pressure producing it. That means 
that in Figs. 35 and 36 the distances OD and GK will represent the 

same interval of time and will denote -— of a second if the fre- 

60 

quency is 60 cycles per 'second. 
The important consideration 
at this point is that the current 
and the pressure do not start out 
from their zero values at the 
same instant. As shown in 
Fig. 36, the current lags behind 
the pressure. Fig. 35; that is, from any chosen vertical datum line 
through the points 0, the alternating current arrives at its maxi- 
mum value later than the pressure arrives at its maximum value. 




Fig. 36. Alternating Current Produced in 
Circuit by Alternating Pressure 



200 



DESIGN OF SMALL TRANSFORMERS 57 

The power resulting from such a pressure and current must be 
found by multiplying corresponding instantaneous values together 
and in some way finding the average power during a definite 
interval, say, during the time occupied by one cycle. 

Power Factor. The average power effect depends upon how 
much the current lags behind the pressure and can always be 
computed when certain conditions are known. The lag of current 
behind pressure is expressed by what is termed the power factor, 
which can be determined very quickly and readily by experiment, 
employing a voltmeter, an ammeter, a wattmeter, and (if the fre- 
quency of the supply pressure is not known or is not constant) a 
frequency indicator. 

The power factor is obtained by dividing the indication of the 
wattmeter by the product of the indication of the voltmeter times 
the indication of the ammeter. 

The frequency should be accurately known at the instant the 
indications of the other instruments are observed, since the value 
of the power factor of any given transformer varies with the varia- 
tion in frequency, so that a transformer designed for a particular 
frequency will not operate at the same efficiency at any other fre- 
quency. The frequency depends upon the speed of the generator 
supplying the input power and can only vary because of a change 
in speed of the supply generator. On any given supply circuit the 
frequency variation will not be very great, so that the change in 
the efficiency of operation of a transformer connected with such a 
supply will not be great. If, however, a transformer designed to 
operate on a 60-cycle circuit is connected with a 25-cycle circuit 
having the same pressure, the transformer will not operate properly 
nor with any but a very low efficiency. 

Relation of Power Factor to Other Values in Circuit. The 
power factor of the primary circuit of a transformer may be 
expressed in terms of the resistance of the primary wire, the pri- 
mary current, and the pressure of the supply mains with which 
the primary is connected. If Ep denotes the pressure of the 
mains, Ip the current in the primary, and Rp the resistance of the 
primary, the power factor (p.f.) may be expressed as follows: ^ 

p ■tipi p 
Ep 



201 



58 DESIGN OF SMALL TRANSFORMERS 

from which it is evident that the power factor increases with the 
increase in current, since the appUed pressure Ep and the resist- 
ance Rp of the primary are constants. This further shows that 
the power factor tends to become unity as the output and the 
input of the transformer increase. As the output increases, the 
input increases, and since the apphed pressure Ep is constant, 
the primary current Ip increases as the load increases. 

For small transformers the power factor is much less than 
unity, while it may be as high as 0.98, or 98 per cent, for very 
large transformers. For a transformer of 2000 watts output the 
power factor at the rated output of 2000 watts will probably not 
be greater than 0.80, or 80 per cent. If W equals the total input 
in watts, the following is true 

TF = Ep7pXp.f. 
which equation may also be written 

W 



h 



EpXp.f. 



Assuming the power factor to be 0.8, the applied pressure 110 
volts, and the input 2500 watts, the equation becomes 

2500 2500 ^o, 

= 28.4 amperes 



" 110X0.8 S^ 

Resistance of Primary Windings. Having found the current 
in the primary windings, if the loss in the primary is known, then 
the resistance of the primary windings can be computed. 

It will now be in order to decide upon the proportion of the 
total loss which shall be assigned to the primary. The total loss 

is to be 2500-2000, or 500 watts, and two-fifths of this, or 500X-|-, 

5 

or- 200 watts, should be assigned as core loss, leaving 300 watts for 

both primary and secondary losses. It is usually 'advisable to 

make the loss in the primary less than the loss in the secondary. 

In the present design the loss in the primary will be assumed to 

be 100 watts, and the loss in the secondary 200 watts. The 

loss in the primary may then be expressed by 

i^p7p2 = 100 



202 



I 



DESIGN OF SMALL TRANSFORMERS 59 

and as Ip has been found to be 28.4 amperes, the resistance of the 

primary will be 

P 100 100 ^,^_ , 

Secondary Current. Since the loss in the secondary has been 
decided upon as 200 watts, the resistance of the secondary wind- 
ings may be found when the value of the current in the secondary 
is found. 

Since the primary current has already been found to be 28.4 
amperes, if the ratio between the primary current and the second- 
ary current is fixed, the value of the secondary current can be 
found. The ratio of transformation of currents in this case, on 
the assumption of no losses and unity power factor, will be exactly 
in inverse value to that of the ratio of the secondary pressure 
to the primary pressure, which is as 20 to 1. Therefore, assuming 
that there are no losses 

Ip 20 

or 

7 1 T 28.4 . , 

20 "^ "" ~20" a^iperes 

Since the transformer does not give out all the input power 
as useful power, a modification of the current ratios must be 
made. Since the ratio of the secondary pressure to the primary 

' pressure has been fixed as 2200 to 110, that is, -tt^, or — — , or ^, 

o 

and since the power output is only — of the power input, the 
following equation is a very close approximation of the relations: 

Ep Ip = -^ Es Is 

o 



Es = 20E 



Eph = ^X20 Epis 
o 



7p=25 7a 
from which 



7s = 777 7p= 7^ = 1.13 amperes 
25 25 



i 



60 DESIGN OF SMALL TRANSFORMERS 

a value somewhat less than the 1.4 amperes found without taking 
the efficiency and power factor into account. 

As the power factor of the secondary of a step-up transformer 
is less than that of the primary, owing to the greater number of 
turns, an assumption of a power factor of 65 per cent, or 0.65, for 
the secondary of the transformer under consideration will not be far 
from ordinary practice. This will make 

Is = -—7- = 1.7 amperes 
0,65 

Resistance of Secondary Windings. Since the secondary loss 
has been assumed to be 200 watts and the secondary current has 
been found to be 1.7 amperes, the resistance of the secondary may 
be found from the equation 
Rs I s^ = 200 

whence is derived 

„ 200 200 nc^ . 

Size of Core. Before proceeding with the computations to 
find the length and size of the primary and the secondary wires, 
it will be necessary to ascertain the width and depth of the iron 
core, in order to know about the length of a single turn of wire 
for primary and secondary. The computation of the core dimen- 
sions will now be considered. 

Mean Length. The core loss will be assumed to be 0.204 
watt per cubic inch, a value found to be proper by computation 
and by actual experiments. The core loss having been assigned 
as 200 watts, if the loss per cubic inch is 0.204 the volume of the 

iron core must be , or 980 cubic inches. If the cross- 

sectional area of the core be taken as 88.8 square centimeters (as 
already assumed), or an equivalent of 13.76 square inches, the 

mean length of the core will be -7-^, or 71 inches. The mean 

13. /o 

length is denoted by the dotted line ABCD in Fig. 37. It has been 

found that better results may be obtained by making the two 

limbs of the core on which the coils are to be wound longer than 

the other two connecting portions, or yokes. A desirable ratio is 



204 



DESIGN OF SMALL TRANSFORMERS 



61 



to make the length of the yoke about two-thirds of the length 

2 
of the wound portion. In the present case AB equals — BC and 

o 



2{AB-\-BC) equals 71 inches and therefore 

2 



2(-^BC+BC) 



71 



from which 



■^BC+2BC 



71 



4-BC^-^ 

3 3 



BC = 71 



^BC = 71 



5(7 = ^X71=21.3 in. 



If BC is 21.3 inches, then 



AB = -jX21.3 



42.6 
3 



14.2 in. 



For practical reasons BC may be called 21 inches and AB 14.5 
inches. As a check on the computations, adding AB to BC and 
multiplying by two should give the 
mean length; and in this case it does 
give 71 inches, the iiiitial value computed. 
Length of Strips. By building the 
iron core of rectangular strips, only two 
sizes will be needed. Fig. 38. The dis- 
tance ah has been decided upon as 14.5 
inches, which will be the length of the 
shorter strips CD and OH, The longer 
strips EF and AB will be 21 inches long, 
as they correspond to the distance he. 
The core is built up to the desired thick- 
ness by the proper number of plates 
placed one above another, so placed as Fig.s?. 
to * 'break joints," or to lap one another 
at the corners. If the first layer is arranged as indicated in Fig. 38, 
then the next layer would be arranged as in Fig. 39, the third 
layer as in Fig. 38, and so on. Enough layers should be added 
to make the thickness the same as the width of each strip. 




Diagram Showing Mean 
Length of Transformer Core 



205 



62 



DESIGN OF SMALL TRANSFORMERS 



Width and Depth of Core. If the cross-section of the iron core 
is to be square, the width and the depth of the section will be the 
same, and if the cross-sectional area of the core is 13. 76 squ are 
inches, the width and depth of the core will each be Vl3.76, or 
3.72 inches. 

The primary wire might be wound on a form having a square 
central hole large enough to allow it to be slipped easily over the 
core, but it is usually easier and more satisfactory to wind the 
primary and also the secondary wire on a cylindrical form. This 
method requires slightly more wire in winding, the additional 



r 



/od- 



Fig. 38. 



Arrangement of First Layer 
of Core 



Fig. 39. 



Arrangement of Second Layer 
of Core 



length being the difference between the length of four sides of a 
given square and the length of a circle that passes through the 
four corners of the square. This is clearly illustrated in Fig. 40. 
It is evident that the length of the circumference abed is more 
than the length of the sides of the square AB-\-BC-\-CD-\-DA. 

Weight of Iron Core. In order to know approximately what 
the cost of the core material will be, it will be necessary to know 
the volume of the core in cubic inches and the weight of the 
iron used per cubic inch, or the volume of the core in cubic feet 
and the weight of the iron in pounds per cubic foot. 



206 



DESIGN OF SMALL TRANSFORMERS 63 

The weight of ordinary iron, such as would be used in the 

transformer described, would be very close to 490 pounds per 

cubic foot. If the weight per cubic inch is desired, it wall be 

490 

-— — , or 0.283 pound since there are 1728 cubic inches in 1 cubic 

foot. Since the volume of the core is 980 cubic inches, the weight 
of the core will be 980X0.283, or 277 pounds. 

Details of Construction of Core. "The eddy-current loss in a 
transformer depends upon the ohmic resistance of the iron of 
which the core is made, but it also depends in a very large degree 
upon the care with which the core is constructed. The strips of 
iron constituting the core should have all burrs and sharp edges 
carefully removed so that it will be impossible for a point on one 
strip to make metallic contact with another strip. If sharp points 
protrude from the corner of a strip, it might make a good elec- 
trical contact with another strip, even through a good coat of 
insulating varnish. If the core were in one solid piece, the 
pressure induced in it owing to the changing magnetic flux would 
during each cycle produce a very large current in the solid iron 
core because of the very low resistance of the solid core due to 
its large cross-section. Great care should be exercised in the 
building of all parts of a transformer, but especial care should 
be taken in the construction of the core. 

Length of Primary Wire. Having determined the cross- 
section of the core and having decided to wind the primary turns 
in circular form, the length of a single turn may be ascertained 
by knowing the diagonal of the square cross-section of core; 
that is, the length of BD in Fig. 40. Of course, the diameter 
of the circular turn of primary wire will need to be slightly 
larger than the diagonal BD in order to be easily slipped on over 
the core. 

The diagonal BD may be found from the well-known rela- 
tion existing between the two sides and the hypotenuse of any 
right-angled triangle — the hypotenuse squared is equal to the sum 
of the squares of the other tico sides. In the case under considera- 
tion, the following equation expresses the relation: 

BD' = BC^+CD'' 



207 



64 



DESIGN OF SMALL TRANSFORMERS 



Substituting the numerical value 3.72 for BC and CD gives 

BD^= 3.722+3.722 

= 13 .76+ 13.76 =27.52 
BD =V27.52=5.2in. 

In order to provide for sufficient clearance, the diameter of a 

single turn of the primary should be made 5.5 inches. 

The length of one turn of primary tD will therefore be 

17 27 
3.1416X5.5, or 17.27 inches, which is equal to , or 1.44 feet. 

If the length of one turn of the primary winding is 1.44 feet and 




4 



Fig. 40. Determining Diameter of Circular Turn of Primary Wire 

there are a total of 154 turns, the total length of the primary 
winding will be 221.76 feet, or practically 222 feet. 

Size of Primary Wire. The size of the primary wire may be 
determined from the well-known relation between the length of 
a round copper wire, its specific resistance, and its cross-sectional 
area, expressed as follows: 



208 



DESIGN OF SMALL TRANSFORMERS 65 

If the value of the cross-sectional area is desired, the expression 
may be arranged as follows: 

In which A denotes the cross-sectional area of a round copper 
wire, - expressed in number of circular mils; p denotes the resist- 
ance, in ohms, of a round copper wire 1 foot long and 1 circular 
mil in cross-section, which at an ordinary temperature, say about 
70° F., may be taken as 10.54 ohms; / denotes the length of the 
given wire expressed in feet; and R denotes the resistance of the 
given wire in ohms. 

The length of the primary wire having been found to be 
222 feet and its resistance to be 0.1239 ohm, the cross-sectional 
area becomes 

. 10.54X222 ^„„„ . ., 

-^^ 0.1239 =^^^^^ '''' "^'^^ 
The size of wire having a cross-section the nearest to this value 
is No. 8 B.&S., which according to column D, Table II, has a 
cross-section of 16,512 circular mils, a value somewhat less than 
that just computed. If No. 8 wire is adopted, the primary loss 
will be slightly in excess of the assumed loss of 100 watts. 

Winding Space Required by Primary Wire. If No. 8 wire 
is used as the primary and it is covered with double-wound cot- 
ton insulation, its over-all diameter will be 0.142 inch. Since 

154 
there are to be — -, or 77 turns per coil on the primary, the 

length required for 77 turns of No. 8, laid side by side, must be 
77X0.142, or 10.9 inches. The length actually needed to accom- 
modate the 77 turns will be somewhat greater, since there are 
always slight bends in such large wire. Probably the space 
needed will be nearer \\\ or 12 inches. By consulting Fig. 38, 
it is evident that there is ample winding space on the longer 
leg of the core. 

Length of Secondary Windings. The length of the secondary 
may be determined when the length of a mean turn is known. 
The secondary windings of a step-up transformer will occupy 
more winding space than is required for the primary windings 
owing to the predominating ratio of insulation to conductor. 



209 



66 



DESIGN OF SMALL TRANSFORMERS 



Considerable space should also be allowed for insulation 
between the primary and the secondary, as indicated in Fig. 41. 
The length of a mean turn may be computed after the diameter 
of a mean turn has been decided upon. It may be found neces- 
sary to make a different assumption as to the length of a mean 
turn after finding the length and the cross-sectional area of the 
secondary wire. The mean diameter of the secondary will be 
assumed to be 7 inches, and the length of a mean turn must be 



/^^(JLAT/o/y 




^J£CO/VDA/^r 



p/^/z^A/^r 



Fig. 41. Cross-Section of Core, Windings, and Insulation 

3.1416X7, or 21.9912 inches, say 22 inches. Since the ratio of 
transformation is 20 to 1 for the transformer being designed, 
there would be 20 times as many turns in the secondary winding 
as there are in the primary windings, were there no losses of any 
kind in the transformer. Since the efficiency is only 80 per cent, 

or — , the ratio of transformation must be multiplied by — in 

order to have the requisite number of turns to maintain a sec- 
ondary terminal pressure of 2200 volts with the secondary output 



210 



DESIGN OF SMALL TRANSFORMERS 67 

equal to 2000 watts. This will mean — -X— Xl54, which equals 

1 8 

25X154, or 3950 turns of secondary winding. The total length 
of the secondary will be 22X3950, or 86,900 inches, which equals 
8^0^ ^^ „241 feet. 

Size of Secondary Wire. Since the length of the secondary 
is 7241 feet and its resistance is 69.2 ohms, its cross-sectional 
area in circular mils is 

' I 10.54X7 241 ^.„^ . ., 

This corresponds very nearly to No. 20 B.&S. gage, according to 
column D, Table II. The over-all diameter of this size of double 
cotton-covered wire is given in column F as 0.041 inch. 

Winding Space Required for Secondary Wire. Since there 
is to be a total of 3950 turns in the complete secondary, if the 
secondary windings are distributed over the two legs of the core, 

there will be — — , or 1975 turns per leg. The winding cross- 

sectional space required by this number of turns will be 0.041 X 

1975X0.041, or 3.31 square inches. If a length of 12 inches 

along the leg be allowed for the secondary, the depth of the wind- 

3 31 . 

ings will be -^—, or 0.27 inch. If the secondary windings are 
\-Z 

extended over only 6 inches of length along the core, the depth 

of the windings must be 0.27X2, or 0.54, or somewhat over \ inch. 

Weight of Secondary Wire.. By consulting column G, 
Table II, it is seen that the weight of No. 20 B.& S. gage, 
double cotton-covered wire, is 3.22 pounds per 1000 feet. The 
weight of the secondary wire for the transformer being designed 
is therefore 7.241X3.22, or 23.3 pounds. It will be advisable 
to have 24 pounds to provide for irregularities and connections. 

Winding Secondary. The secondary for a transformer hav- 
ing a secondary pressure of 2200 volts will heed to be wound in 
a number of separate coils or forms. 

Permanent forms made of soft pine wood, well soaked in 
hot paraffin, are the best type to employ. Forms of such wood 



211 






68 DESIGN OF SMALL TRANSFORMERS 

may be readily turned out in a wood-turning lathe or may be 
built up by sawing the sides from thin pine and gluing them to 
a stiff cardboard cylinder. 

For the best insulation the secondary should be wound in I 
sixteen different coils, that is, eight coils per leg. By consulting 
Figs. 38 and 41, the size of the forms for the secondary coils may 
be computed. 

Insulation between Primary and Secondary. The secondary 
windings should be insulated from the primary by means of a 
thick tube of some good insulating material that should extend 
beyond the windings for an inch or more at either end. 

An insulating tube of good quality may be made by winding 
thin cardboard about a cylindrical form of the proper size, gluing 
the layers together and building the thickness of the tube up 
to about I inch. Such a tube may be soaked for a time in hot 
paraffin and then allowed to cool slowly, or it may be carefully 
varnished with a good grade of varnish, allowing the varnish to 
become dry and hard before operating the transformer. 

In winding the fine wire on the forms for the secondary, it 
should not be wound too tightly, which is liable to happen if the 
coils are wound in a lathe. A loose winding allows a free flow 
of oil between the windings, if oil insulation is employed, and 
admits more insulating material if a solid insulator such as para- 
ffin is used. 

Summary. It will be instructive to arrange the important 
results of the design just completed, for comprehensive study 
and comparison, as follows: 

Specifications 

Useful output 2 kilowatts, or 2000 watts 

Frequency 60 cycles 

Efficiency 80 per cent 

Input 2500 watts 

, Primary 

Primary turns 154, total 

Primary current 28.4 amperes 

Primary loss 100 watts 

Primary resistance 0.1239 ohm 

Length of primary wire . . . 222 feet 

Size of primary wire No. 8 B.& S. gage 



212 



DESIGN OF SMALL TRANSFORMERS 69 

Secondary 

Secondary loss 200 watts 

Secondary current 1.7 amperes 

Secondary resistance 69.2 ohms 

Secondary turns 3950, total 

Length of secondary wire. 7241 feet 

Size of secondary wire. . . .No. 20 B.& S. gage 
Core 

Volume of core 980 cubic inches 

Core loss 200 watts 

DESIGN OF I0=VOLT TRANSFORMER 
Specifications. It will be assumed that the pressure is 110 
volts between the wires of the mains supplying power to the 
transformer designed to deliver f kilowatt, or 750 watts, at 10 
volts. The frequency of the supply pressure will be assumed to 
be 60 cycles. 

Ratio of Transformation. If the pressure is to be stepped 
down from 110 to 10 volts, the ratio of transformation will be 

Ep 110 11 
or there will need to be at least eleven times as many primary 
turns as secondary turns. 

Number of Turns in Primary Coils. The necessary number 
of turns in the windings of the primary coils may be found from 
the relation between the primary applied pressure, the cross- 
sectional area of the iron core, the frequency of the applied pres- 
sure, and the density of the magnetic flux in the iron core. The 
relation may be expressed by symbols as follows: 

27r/yly§ fA/^ 

Substituting in the foregoing equation the assumed numerical 
values for the applied primary pressure Ep and the frequency /, 
110 volts and 60 cycles, respectively, the equation will become 
2 2504456X110 ^ 22 504 456X11 

mxAx/^ ~ 6x/ixy^ 

It is now evident that there are two unknowns in the equation, 
that is, the cross-sectional area A of the iron core and the number 
of lines of force /B per square centimeter of cross-section of core. 
The assumptions as to these two quantities must depend some- 



213 



Np = 



70 DESIGN OF SMALL TRANSFORMERS 

what upon the result of experiments with the magnetization of 
iron cores as well as upon experience in designing transformers. 

Under the conditions of operation of an ordinary transformer 
that is built for purposes of experimentation, a fair assumption 
for /3 will be 5000 lines of force per square centimeter. For the 
cross-sectional area A of the core a value of 40.31 square centi- 
meters, or an equivalent of GJ square inches, will be assumed. 

The number of turns in the primary may now be found 

,, 22504456X110 22504456X11 ^^, ^ 

JY = — = = ^04 turns 

'' 60X40.31X5000 6X40.31X5000 



It is evident that if less than 5000 gausses be assumed for M, 

more primary turns will be required. 

Efficiency of Transformer. At this point it will be advisable 

to determine what the efficiency of the proposed transformer 

shall be and the distribution of the various losses. The efficiency 

when the transformer is giving its rated output of 750 watts will 

85 
be assumed to be 85 per cent. This means that only — — of the 

electrical power input to the primary of the transformer will be 
available for useful output from the secondary. If the efficiency 

is — , or 0.85, the input must be —— of the useful output. In 
100 85 ^ 

this case the input will be -^^X750, or 882 watts (very nearly). 

oO 

Distribution of Losses. The losses in electrical power in the 
transformer will total 882 — 750, or 132 watts. The losses may 
be divided into the primary loss, the secondary loss, and the 
core loss. AYhile the primary and the secondary losses are so 
called RP power losses, the core loss may be separated into two 
parts: the eddy-current loss, due to electric currents induced in 
the core; and the hysteresis loss, a heat loss resulting from the 
rapid reversals of the molecules constituting the iron of the core 
caused by the rapid alternations of the primary current. The losses 
in the transformer under consideration will be apportioned as follows : 

Primary loss : 33 watts 

Secondary loss 66 watts 

Core loss 33 watts 

Total loss 132 watts 



214 



I 



j^ DESIGN OF SMALL TRANSFORMERS 71 

V 

Volume of Iron Core. The core must consist of a certain 
volume of iron, and the heating of this whole bulk is represented 
by the so-called core loss of 33 watts. It is obvious that if this 
total core loss be divided by the volume of the core in cubic 
inches, the result will be the core loss in watts per cubic inch. 
As a result of very careful and extensive experiments the follow- 
ing expression represents very closely the core loss in any trans- 
former in watts per cubic inch. 



iKJ-^xC^+a-mifM-) 



10^ V 10' 

After substituting in this equation the numerical values of / and 
/3 and for b a value of 15 (which means the thickness in mils of 
the iron strips of which the core is built up), the following 
equation results: 

PF,=16^xr?25><6^^<«MV(o.0021X60x5000>-»)] 
10^ L 10 J 

The numerical value of 5000^^ is obtained most easily by the 
use of logarithms as follows: 

log 5000i« = log 5000X1.6 

= 3.6989870 X 1 .6 = 5.9183520 

The number corresponding to this logarithm is found to be 
828,600. The equation therefore becomes 



_16,38_ 
' 10 000 000 



"?2^f^|?^0+(0.0021 X60X828600) 



= 0.000001638X[(22.5X36X25) + (0.0021X60X828600)] 
= 0.000001638 X (20250 + 1 044 036) 
= 0.204 watt per cu. in. 

It is now in order to ascertain the volume of the iron core 

in cubic inches. If the loss per cubic inch is 0.204 watt and the 

total core loss is to be 33 watts, the total number of cubic inches 

33 
in the core will be , or 163.2 cubic inches. 

U.<uU'± 

Length of Strips. If the cross-sectional area of the core is 
to be 6.25 square inches, the mean length of the iron core will be 



215 



72 



DESIGN OF SMALL TRANSFORMERS 



163.2 



, or 26.4 inches (very nearly). One-half of this mean length 



IS 



or 13.2 inches. Calling the mean length an even 13 

The width of 



6.25 
26^ 
2 ' 
inches, the size of the iron strips may be found, 
the strips is 2 J inches. 

Suppose the distance ABDC in Fig. 42 denotes the mean 
length of the core, and let the assumption be made that the 
distance AB shall be two-thirds of the distance AC. The following 

relation must then be true: 
AB+AC = 13 
But by the assumption that ^45 equals ^A C 
iAC-\-AC = lS 

3 
3X13 39 
5 " 5 




-AC = 13 



AC 



7.8 in. 



Fig. 42. 



Determining Length of 
Core Strips 



If AB-\-AC equals 13 inches and AC 
equals 7.8 inches then 

^5 = 13-7.8 = 5.2 in. 

To check this result, add 5.2 and 7.8, 

which gives 13 inches. The shorter strips 

may be made 5 inches long and the 

longer strips 8 inches, as indicated hy AB and CD, Fig. 43. 

The volume of the core, using strips 5 inches and 8 inches 
long, may be found by multiplying 5 X (2^)^X2 and adding the 
result to 8X(2i)2x2. This gives 

5 



Volume 



+ 



=[ 



5X(| 



8X^X2' 



X2 



+ 



8X 



250 400 

4 "^ 4 



2 

650 
4 



X2 



5X^X2 



] 



= ^ = 162.5 cu. in. 



which is very close to 163 cubic inches. 

Weight of Iron Core. The weight of the necessary iron for 
the core may be found by multiplying the total cubic inches in 
the core by the weight of 1 cubic inch of the iron that is employed. 
Ordinary soft steel that is used for transformer plates weighs very 
close to 0.28 pound per cubic inch. On this basis the weight of 
the core being considered will be 163X0.28, or 45.6 pounds. 



216 



DESIGN OF SMALL TRANSFORMERS 



73 



I 

^ Resistance and Current of Primary. The dimensions of the 
iron circuit having been determined, calculations pertaining to 
the copper circuits will now be considered. Logically the pri- 
mary circuit will be considered first. Since the power loss in the 
primary was assumed to be 33 watts, the relation between this 
loss, the primary circuit, and the resistance of the primary may 
be expressed by 

Rplp^ = 33 watts 

The next step will be to ascertain if possible the value of 
the primary current Ip in amperes. In order to do this, it will 
be necessary to revert to the 
power input to the primary and 
to the pressure applied to the 
primary. The power input was 
found to be 882 watts. If it 
be assumed that the power factor 
of the primary is unity, which 
will be very nearly the case when 
the transformer is giving its 
normal output, then it may be 
said that the product of the 
primary applied pressure and the 
current in the primary will be 
equal to the power input. Then 
the current in the primary can 
be found by dividing the power 
input by the primary applied pressure as follows: 

882 




Fig. 43. Arrangement of Assembled 
Strips 



h 



110 



= 8 amperes 



From this value of primary current the resistance of the 
primary may be found from 

Rplp^ = 33 or 
: RpXS' = 33 

33 33 
64 



iJp=^=^ = 0.5156ohm 



Length of Primary Wire. Having found the resistance of 
the primary, its cross-sectional area can be computed, if its length 



217 



74 



DESIGN OF SMALL TRANSFORMERS 



in feet can be ascertained. One method of finding the length 
of the primary would be to find the average, or mean, length of 
turn. If the wire is wound in a coil one layer over another, the 
length of a single turn on an outside layer will be greater than 
the length of a turn on an inside layer. The greater the cross- 
sectional area of the wire the greater w411 be the difference between 
the lengths of the outer and the inner turn. 

So far as the amount of wire is concerned, it is advisable to 
make the diameter of the coil as small as possible. To ascer- 







Fig. 44. Determining Length of Primary Wire 

tain the possible minimum diameter that can be slipped over the, 
iron core, it will be necessary to ascertain the length of the diag- 
onal of the square section of the core. The core being of square 
cross-section, 2 J inches on a side, as indicated hy AB and BC, 
Fig. 44, the length of the diagonal ^C is expressed by 



. ^C=1.414X^5 = L414x2i = 1.414x|- = 0.707x5 = 3.53 in. 

Li 

Therefore the smallest circle, or turn of wire, that can be slipped 



218 



DESIGN OF SMALL TRANSFORMERS 75 

over the core will be one whose diameter is not less than 3.5 
inches. To allow for irregularities, the diameter should be about 
3.7 inches, and if the primary coil is wound on a form, the diam- 
eter of the coil will probably be 4 inches. 

If the diameter is assumed to be 4 inches, the length of one 
turn, or ttD, must be 3.1416X4, or 12.5664 inches, which for 
all practical purposes may be taken as 12^ inches. If there are 
to be 204 turns in the primary windings, the total length of the 

2550 
primary will be 204X12^, or 2550 inches, which equals — r^, 

or 212 feet. 

If it should be found necessary to wind the primary in more 
than one layer, it is evident the result just obtained will then not 
be exact. Another calculation could be made to ascertain just 
what the diameter of the inner layer should be in order to wind 
a given length of wire in the required number of turns. There 
will not be much difference between the lengths of the inner and 
the outer turns if there are only two layers of the necessary size 
of wire for the transformer being designed. 

Size of Primary Wire. The length of the primary being 
assumed to be 212 feet, and its resistance Rp being, as previously 
found, 0.5156 ohm, the cross-sectional area of the primary wire 
may be found in circular mils from the fundamental relation 
between the resistance of a round wire, the specific resistance of its 
material, its length, and its cross-sectional area in circular mils. 
This is expressed in symbols by 

in which p denotes the so-called specific resistance of the material, 
that is, the resistance, in ohms, of a mil-foot, a mil-foot being 

inch in diameter and 1 foot long. Since the specific resist- 

1000 

ance of copper wire at ordinary temperatures may be taken as 

10.54 ohms, p equals 10.54 ohms. The symbol I denotes the 

length, in feet, of the round wire being considered, A denotes the 

cross-sectional area, in circular mils, of the wire, and R denotes 

its resistance. 



219 



76 



DESIGN OF SMALL TRANSFORMERS 



As the area of the cross-section of the primary wire is desired, 
the fundamental equation may be solved for A giving 

and, making the proper numerical substitution for symbols, the 
result is 

10.54X212 



A = 



0.5156 



= 4333 cir. mils 



If the diameter of any' round wire is expressed in mils, that 
is, in thousandths of an inch, and the number squared, the result 
will be the number of circular mils in the cross-section of the wire, 
and if the diameter of the wire is desired, it may be found by 
taking the square root of the area, in this case 4333. This is 
65.8 very nearly; that is, the diameter is 65.8 mils, or 0.0658 inch. 

This corresponds very nearly to No. 14 wire, coming between 
No. 14 and No. 12, according to Table II. If No. 12 wire 
should be used, since it is larger than No. 14, and the required 
length is 212 feet, the primary loss will be less than the assumed 
value of 33 watts. This will mean that the efficiency of the 
finished transformer will be slightly greater than the initial 
assumption of 85 per cent. 

Weight of Primary Wire. If No. 12 wire is used, the weight 
of double cotton-covered wire will be found from Table II, 
column G, to be 20.1 pounds per 1000 feet. The weight of 1 
foot will therefore be 0.0201 pound, and the weight of 212 feet 
will be 212X0.0201, or 4.26 pounds. 

Space Required by Primary Windings. It will be important 
to know approximately the amount of space required for both 
the primary and the secondary windings, in order to be sure the 
opening through the transformer is of sufficient size to accommo- 
date the windings. A method for finding the winding space will 
be outlined. If the total number of turns in a coil is known 
and the cross-section of each wire is known, it is evident that 
the area occupied by all the turns will equal the product of the 
number of turns times the area required by each turn. If the 
wire is round, it requires a square area having each side equal 
to its diameter. For the primary being considered, if it be 



220 



DESIGN OF SMALL TRANSFORMERS 77 

204 
divided into two coils, there will be — -, or 102 turns per coil. 

Zi 

The diameter of double cotton-covered No. 12 wire is seen in 

column F^ Table II, to be 0.092 inch, and therefore each wire 

will require an area equal to 0.0922. The area required by 102 

turns will be 102x0.092^, or 0.8633 square inch. Another method 

of considering the matter is to ascertain the necessary length to 

accommodate 102 turns of this wire if in one layer. This is 

obviously 102X0.092, or 9.38 inches. Since the length of the 

longer leg of the iron core of the transformer is only 8— 2 J, or 

5J inches, Fig. 43, it is evident that the primary will need to be 

wound in two layers. The length of winding space required for 

9 38 
a two-layer coil will be -^, or 4.69 inches. Since the wire can- 

not be wound without some irregularities or slight bends, it 

always occupies slightly more space than that figured from its 

diameter and the number of turns. It will be well to allow 5 

inches in this case instead of 4.69. 

If the winding space is assumed to be 5 inches, that will 

leave 5.5 — 5, or 0.5 inch between the ends of the coil and the 

0.5 
iron yoke of the core. There will be -^, or 0.25 inch between 

either end of the coil and the core. This will allow the use of 
soft pine blocks or boards previously soaked in hot paraffin and 
placed between the ends of the coil and the iron core. 

Number of Turns on Secondary. Having settled the data for 
the core and the primary, the next in order is the secondary. 
Since the ratio of transformation for the transformer was decided 
to be 1:11, if there were no losses in the transformer, there would 
be 11 times as many secondary turns as there are primary turns. 
However, since the efiiciency of the transformer is only 85 per 

cent, if the secondary current is maintained at — of the primary 

current, the ratio of secondary turns to primary turns cannot be 

— but will be nearer — — — — — , or jr^', that is, there will be 
11 110X0. oo c/o.o 

— - times as many turns on the secondary as on the primary. 



221 



78 DESIGN OF SMALL TRANSFORMERS 

100X204 
The number of turns on the secondary is therefore — — -^ — , or 

935 

21 turns. If the secondary is wound in two coils, each would 

have 11 turns, making a total of 22 turns for the complete 

secondary. 

Length of Secondary Wire. The size of the wire for the 

secondary may be found in a manner similar to that adopted in 

finding the size of the primary. If the length of the secondary 

can be ascertained, it is evident that the cross-sectional area As 

may be found from the expression 

lis 

In order to find h, the length of a mean turn of the secondary 
may be assumed by consulting Figs. 43 and 44. From the data 
given in Fig. 44, the length of the diagonal AC was found to 
be about 3.5 inches. The diameter of the primary coil was 
assumed to be 4 inches, and the diameter of the secondary will 
be assumed to be 5 inches. The length of the turn will be 
3.1416X5, or 15.7080 inches, or 1.3 feet. The. length of the sec- 
ondary is 21X1.3, or 27.3 feet. 

Size of Secondary. The power loss in the secondary was 
assumed to be 66 watts, which may be written 

i^.7/ = 66 
Since the current in the secondary is 11X8, or 88 amperes, the 
resistance of the secondary becomes 

i^,=^ = -M^ = 0.008522 ohm 
88^ /744 

By making the proper numerical substitutions in the fundamental 
area equation given, the following equation is obtained: 

. 10.54X27.3 287.84 „_„„ . ., 
^^= 0.008522 =0008522 = ^^^^^ '''' ^^^^ 

The nearest wire in Table II is found in column D to be 
No. 6. Since the sectional area of No. 6 wire is somewhat less 
than the computed value required, if a No. 6 is used, the effi- 
ciency of the transformer will be slightly less than 85 per cent. 
The increased efficiency due to the larger size adopted for the 



222 



DESIGN OF SMALL TRANSFORMERS 79 

primary will be offset by the lessened efficiency due to the smaller 
size of wire for the secondary. A No. 12 wire can be wound on 
close to (in parallel with) the No. 6, and the area of the two 
wires will be very nearly equal to the computed value. 

In order to avoid labor and difficulty in winding due to the 
stiffness of such a large wire as a No. 6, two smaller wires hav- 
ing a combined area equal to 33,893 circular mils may be used. 
If two wires are used, of course, each wire must have an area of 

— - — , or 16,946 circular mils. Two No. 8 wires or five No. 12 

wires might be used. In the latter case the total sectional area 
will be 5X6528, or 32,640 circular mils, which is slightly less than 
the computed area. If desired, eight No. 14 wires might be used, 
the combined area of the eight wires being 8X4106, or 32,848 
circular mils, which is very close to the desired value. 

Connecting Coils for Different Pressures. It may be observed 
that the primary, being wound in two coils, may be connected 
with a 55-volt instead of a 110- volt circuit, if the two primary 
coils are connected properly in parallel. 

In the same manner, if the two secondary coils are connected 
together in parallel instead of in series, the secondary pressure 
will be only 5 volts instead of 10 volts. However, in this case 
the secondary current will be 88X2, or 176 amperes, instead of 
88 amperes. This is, of course, assuming that the frequency is 
60 cycles. 

If each secondary coil is connected up separately with an 
operating load, each circuit will have a pressure of 5 volts and a 
current of 88 amperes. This is the same as stating that if the 
total useful output is 750 watts when both coils are acting to- 
gether in series, the useful output per coil when acting separately 

750 
is — -, or 375 watts. Each secondary coil gives a useful power 

output of 375 watts irrespective of the way it is connected. 

Pressure Drop in Transformers. When there is no load on 
the secondary of a transformer and its primary is connected with 
the supply mains, a certain definite pressure exists between the 
terminals of the secondary. This is called the secondary electro- 
motive force. If a load is connected with the secondary, the 



223 



80 DESIGN OF SMALL TRANSFORMERS 

pressure between the secondary terminals is less than when on 
open circuit, being less by the pressure drop, in volts, in the 
secondary windings. This pressure drop is proportional to the 
amount of current in the secondary. Since the resistance of the 
secondary is constant, the Rsis pressure drop in the secondary 
wire will vary with the current, that is with the secondary load. 
If E' denotes the total induced pressure in the secondary, or 
the secondary terminal pressure at no load, then the terminal 
pressure E at any load current 7s will be very nearly 

E= E' — Rs Is 

The induced pressure E' is very nearly constant, and, Rs being 
constant, the operating terminal pressure E decreases as the useful 
output 7s increases. 

Winding to Produce Different Pressures. As it is sometimes 
desired to obtain a variety of secondary pressures from the same 
transformers for experimental purposes, a method of so winding 
the 110-10 volt transformer just described as to make this pos- 
sible will be outlined. 

Since there are 22 turns of wire on the secondary, if the 
secondary pressure is 10 volts, the induced pressure per turn is 

— , or 0.454 volt. If two turns were^ wound on the transformer, 

the pressure between the terminals of the two turns would be 
2X0.454, or 0.908 volt. Let it be supposed that one of the 
secondary coils is wound in two sections, one section having four 
turns and the other section having seven turns. The pressure of 
one section will be very nearly 0.454x4, or 1.816 volts, and the 
pressure of the other will be very nearly 0.454x7, or 3.178 volts. 
If the 1.816- volt section be connected in series with the secondary 
coil containing 11 turns, which has a pressure of 5 volts, the 
pressure of the combination will be 1.816+5, or 6.81 volts. If 
the 3.17-volt section be connected in series with the 5- volt coil, 
the combined pressure will be 3.17+5, or 8.17 volts. 

Connecting Coils in Opposition. It is an interesting and 
valuable fact that the coils of a transformer may be connected 
together in opposition, as though they w^ere cells. This can easily 
be tried out in practice with coils wound as suggested above. 



224 



DESIGN OF SMALL TRANSFORMERS 



81 



Let it be supposed that the L816-volt coil is connected in 
opposition with the 3. 178- volt coil. In this case the resulting 
pressure will be 3.178-1.816, or 1.362 volts. If the 3.178-volt 
coil is connected in opposition with the 5-volt coil, the resulting 
pressure is 5—3.178, or 1.82 volts. It is thus evident that a large 
number of different pressures may be obtained. 

These pressures might be tabulated as follows: a negative 
sign indicates that the coils are connected together in opposition, 
while a positive sign indicates that they are connected properly in 

series* 

1.816+5 = 6.816 volts 

3.178+5 = 8.178 volts 

5-1.816 = 3.184 volts 

5-3.178 = 1.821 volts 
5+3.178-1.816 = 6.362 volts 
5 + 1.821-3.178 = 3.642 volts 

By dividing the second secondary coil in a slightly different 
manner, a larger number of variations could be obtained. 

Autotransformers. The manner of connecting coils together 
as outlined above indicates the fundamental principles of the so- 




Fig. 45. Connections for Boosting and Crushing Autotransformer 

called autotransformer. This is a transformer so constructed and 
arranged as to increase slightly the pressure between two mains so 
as to compensate for an increased line drop due to an increase in 
line current corresponding with increased load. 

The principle of a "boosting'* autotransformer is illustrated in 
Fig. 45. One set of coils is of large wire, through which the main 
line current passes. If the load is increased by connecting in 
more lamps, the magnetizing effect of the series turn is increased, 
increasmg the magnetic flux in the core and thus increasing the 



225 



82 DESIGN OF SMALL TRANSFORMERS 

pressure of the second set of coils, which may be considered as a 
secondary. As the terminals of the secondary are connected with 
the line wires in parallel with the load, the applied pressure at 
the load is increased. 

If the terminals should be reversed between 1 and 2, Fig. 45, 
the autotransformer w^ill act as a ''crusher" instead of as a 
booster. 

DESIGN OF 22=VOLT TRANSFORMER 

Specifications. Having considered in some detail the design 
of several transformers for different pressures and outputs, the 
outline of the design of a step-down transformer from 220 to 
22 volts, giving a useful output of 2J kilowatts, or 2500 watts, 
will be more briefly presented. The frequency of the pressure and 
current will be assumed to be 60 cycles, and the efficiency of the 
transformer will be assumed to be 83J per cent. 

Basic Values. If the output is 2500 watts and the efficiency 
is 83 J per cent, or 0.83|, the input must be as follows: 

T , 2500X100 2500X100 750000 ^^^^ ,, 
^"^^'==-^^ = -250— = ^5^ = ^^^^ ^"''^ 

3 

If the pressure is to be stepped down from 220 to 22, the ratio 

22 1 

of transformation is — — , or — . Assuming the power factor to be 

unity, the current in the primary will be -^^, or 13.6 amperes. 

If a power factor of 90 per cent is assumed, the primary current 

win be -T^r, or 15.1 amperes, 
u.y 

If the primary current is to be 15.1 amperes and the efficiency is 

to be 83J per cent, the secondary current will be 

/. = 15.1X10X^^L. = 151X^ 

15100X3 45300 ,_„ 
= ^60- = ^50-='^''"'°P- 

If the efficiency were 100 per cent, since the secondary pressure is 



226 



DESIGN OF SMALL TRANSFORMERS 83 

only — as miTch as the primary, the secondary current would 

need to be 10 times as great as the primary current. Because the 
output is only 83J per cent of the input, the secondary current 
will need to be larger than ten times the primary current. 

Distribution of Losses. If the useful output of the trans- 
former is 2500 watts and the input is 3000 watts, the losses are 
3000—2500, or 500 watts. The total loss may be apportioned as 
follows : 

Primary loss 150 watts 

Secondary loss 200 watts 

Core loss „ 150 watts 

Total loss 500 watts 

Volume and Weight of Core. If the total core loss is 150 

watts and it is assumed that the loss is 0.204 watt per cubic inch of 

150 
core iron, then the volume of the iron core will be , or 735 cubic 

inches. If the iron weighs 0.28 pound per cubic inch, the total 

weight of the core will be 735X0.28, or 205.8 pounds. 

Mean Length of Iron Core. Since the volume of the iron 

core is to be 735 cubic inches and its cross-section is to be 9 

735 
square inches, its mean length will be —— , or 81 inches. If the 

y 

shorter leg or yoke is to be two-thirds of the longer leg, then 

AB equals i AC, Fig. 42. But by computation 

AB+AC = ^ in. 
and since AB equals i AC, 

f AC+AC = i0.5 in. or 
I ^C = 40.5 in. 



From which 



^^^3X405^121^^24.3 in. 
5 5 



If ^C equals 24.3 inches, and AB equals i AC, 

AB = 24:.3X^=^ = IQ.2 in. 
o o 



227 



84 DESIGN OF SMALL TRANSFORMERS 

Cross=Sectional Area of Iron Core. The cross-sectional area! 
of the iron core will be assumed to be 58.05 square centimeters, 

which corresponds to ' , or 9 square inches, and if the cross- 

D.40 

section is made square, the section will be 3X3 inches. 

Number of Turns in Primary Windings. The necessary 
number of turns in the primary windings are found from 

^ _ WXE^ 
" V2TrfAM 
Assuming M to be 5000 and substituting values throughout, the 
equation becomes 

,. _ 220X100 000 000 

IS p 




L414X3.1416X60X58.05X5000 

■ 2200000 

L414X3.1416X6X58.05X5 

^ 220000 

L414X3.1416X58.05X3 
= 284 turns 

284 
which will allow -— , or 142 turns for each of two primary coils. 

Length of Primary Wire. If the section of the core is square 
and 3 inches on a side, one turn of primary will be about 14.14 
inches in length. This will make the total length of the primary wire 
equal to 284x14.14, or 334.6 feet, which may be taken as 335 feet. 

Resistance of Primary. If the power loss in the primary is 
to be 150 watts and the primary current is 15.1 amperes, the 
resistance of the primary is found from 

which, on solving for i^, gives 

i^p = i^ = 0.6576 ohm 

228 
s 

Size of Primary Wire. If the resistance of the primary is 
0.6576, the cross-sectional area of the wire may be found as 
follows : 

. I 10.54X335 ._^ . .. 
"^^'-R^ 0.6576 =^^^^ '''' ^^^ 
The wire nearest to this is a No. 12 B.&S. 



228 



DESIGN OF SMALL [TRANSFORMERS 85 

Number of Turns in Secondary. If there were no losses in 
the transformer, the ratio of turns would be the same as the 
ratio of pressures; but on account of losses the secondary turns 
will need to be somewhat greater than this ratio, as indicated 
below. 

Es^ 1 
E^ 10 



but 



_N^ 1 



10X0.831 






IX 100 


IX 100 


10X83 J 


lOX 


250 
3 


300 

= 


30 


3 

= — 



10X250 250 25 



or 



and since the turns on the primary have been found to be 284, 
the number of turns on the secondary will be 

i\r,=Ax284 = ?^ = 34 turns 

.34 
This allows — , or 17 turns for each of two secondary coils. 

Resistance of Secondary. If the secondary is to be designed 
in accordance with the assumed secondary power loss, then 

Rsis' = 200 

Since h has been found to be 179 amperes 

„ 200 200 ^^^.^ , 
^^=l79i = 3204r = '-'^'' ^^" 

Size of Secondary. Since the transformer is a low-pressure 
transformer, the insulation of the secondary will not require so 
much space as would be necessary on a high-pressure winding. 
Also the insulation between the primary and the secondary will 
not need so much space as is necessary on a high-pressure trans- 
former. 



229 



86 DESIGN OF SMALL TRANSFORMERS 

The diameter of the primary is assumed to be about 4.5 
inches, and if the secondary is assumed to be 5.5 inches in diam- 
eter, this will allow sufhcient space for insulation between the 
primary and the secondary. 

If the diameter is 5.5 inches, the length of one turn will be 
about 17.3 inches, and as there are to be 34 turns on the second- 
ary, the length of the secondary will be 17.3X34, or 588.2 inches, 

which equals , or 49 feet. 

The cross-sectional area of the secondary will therefore be 

. 10.54X49 c^^^^ . ., 

Relation between Circular and Square Mils. It is sometimes 
desirable in constructing a transformer having a large secondary 
wire, demanded because of large power output and low secondary 
pressure, to economize space in winding by having the large wire 
of rectangular cross-section. This necessitates some knowledge 
regarding the relation between the area of a circular mil and the 
area of a square mil. Since the area of any circle is JttD^, or 
0.7854 1)2, if D denotes the diameter of the circle, the area of a 
square that will just include the circle will be expressed by Z)^. 
It is therefore evident that the ratio of the area of the square to 
the area of the inscribed circle may be expressed as follows: 

Area of square _ Z)^ _ 1 



Area of circle 0.7854 D^ 0.7854 
or 

Area of circle = 0.7854 X area of square 

This result may be applied in practice as follows: Suppose the 
area of a square is given in square mils. The area may, for 
example, be 625 square mils. The length of each side of the 
square is 25 mils since 25X25 equals 625. It is now desired to 
know the area of this square in circular mils, which is 625 divided 
by 0.7854, or 795 circular mils. Whenever the area of a wire 
is given in circular mils, its equivalent area expressed in square 
mils may be found by multiplying the number of circular mils 
by 0.7854. 



230 



DESIGN OF SMALL TRANSFORMERS 87 

Use of Copper Strips. The application of this principle to 
the secondary of the transformer being considered is important. 
The area of the section of the secondary has been found to 
be 83,300 circular mils. The number of square mils will be 
83300X0.7854, or 65,424 square mils. It will now be supposed 
that the secondary wire is to be of copper strip J inch in thick- 
ness. Knowing its area in square mils to be 65424 and its 
thickness to be f inch, which equals 0.125 inch, or 125 mils, the 

65424 
width of the copper strip must be , or 523 mils, which 

523 
equals rp^j or 0.52 inches. Copper strip J inch thick and about 

J inch wide may be used for the secondary. 

Since there is a length on the core for winding of 24—2, or 22 
inches, if copper strip ^V inch thick and 2to inches wide is used, 
the secondary will have to be wound in two layers. Copper strip 
Y6 inch thick and 1 inch wide might be used, and this could be 
wound in one layer. The secondary might consist of four strips 
each -^ inch thick and 1 inch wide, forming a laminated con- 
ductor. In this form the strips would not be insulated from each 
other but would be placed one on top of the other and then 
wound over with insulating tape or cotton cloth well shellacked 
after being wound over the copper strips. 

DESIGN OF CASE FOR 220=VOLT TRANSFORMER 

Uses of Oil. As has been stated, oil is used for the purpose 
of insulation, and it also conducts away from the copper windings 
and the iron core the heat produced there. The oil serves as a 
vehicle for carrying away the heat by so-called convection cur- 
rents, or circulation in the oil, based upon the principle that heat- 
ing any portion of a fluid makes its weight less than that of 
the colder portion and causes it to rise. After it has risen to the 
surface of the fluid, it becomes cooled and then sinks to the 
bottom. This phenomenon constitutes the so-called convection 
currents. The surface of oil has a very important characteristic 
of radiating heat very rapidly. For example, if the temperature 
of the oil is 18 degrees higher than the surrounding air, then 
29.68 British thermal heat units are given off from 1 square foot 



231 



88 DESIGN OF SMALL TRANSFORMERS 

of oil surface every hour. One British thermal heat unit, or 
B.t.u., means, the amount of heat necessary to raise the tempera- 
ture of 1 pound of water 1° F. This is equivalent to 778 foot 
pounds of work. Therefore 29.68 B.t.u. are equivalent to 0.011 
horsepower, or 8.2 watts. If the total area of oil from which 






Fig. 46. Framework for Installing Transformer in Oil 

radiation may take place is, say, 6 square feet, then power may 
be radiated at the rate of 8.2x6, or 49.2 watts. 

A transformer designed for a certain output as an air-cooled 
transformer will have an increased output when immersed in oil. 

Case for Oil Insulation. An arrangement for installing a 
220- volt transformer in oil is shown in Figs. 46 and 47. A wooden 
framework is first constructed as shown in Fig. 46. Soft pine is 
a very desirable wood for the purpose and may be about 2^ or 



232 



DESIGN OF SMALL TRANSFORMERS 



89 



3 inches wide by | inch thick. The framework may be provided 
with cross strips across the bottom side strips, properly spaced to 
allow the iron core of the transformer to rest on them. The core 
may be prevented from sliding about by small blocks properly 
arranged and screwed to the cross strips. It will be advisable to 
assemble the framework with brass screws. 

The framework may be covered with ordinary roofing tin, as 
shown in Fig. 47, and all joints carefully soldered to make the 




Fig. 47. Completed Oil-Case for Transformer 

metal box oil-tight. The wooden framework makes the case stiff, 
and the thin roofing tin provides a light covering. 

A faucet may be soldered into one side near the bottom to 
facilitate the removal of the oil when necessary. 

The primary and secondary terminals may be held in place 
by being passed through porcelain tubes inserted in holes in a 
strip properly arranged across the top of the wooden framework. 



233 



90 DESIGN OF SMALL TRANSFORMERS 

RECAPITULATION 

Most Important Factors in Transformer Design. Having 
considered the general principles governing the design and opera- 
tion of both step-up and step-down transformers and having in 
addition considered the design of transformers having specified 
characteristics, it will be well to briefly reconsider the more 
important factors entering into the design of transformers. 

The first consideration in the design of a transformer may be 
either secondary pressure or useful output, but it is usually both. 
The primary pressure and frequency are usually fixed by the 
character of the system supplying the power for the primary. 

After the secondary pressure and the useful output comes the 
question of the efficiency. Any assumption may be made as to 
efficiency, but whatever assumption is made, the results will of 
course depend upon the assumption. In general, if a high effi- 
ciency is assumed, the first cost of the transformer will be high 
and the cost of its operation will be low; that is, the interest on 
the investment will be large and the cost of operating power will 
be low. A proper balance between these two values is of course 
desirable. For a transformer that is to be operated continuously 
day and night at normal rated useful output, the first cost should 
be relatively large and its losses small. On the other hand, a 
transformer that is to be used intermittently for experimental 
purposes should cost less and be less efficient; that is, have 
greater losses when operating at full-load output. 

Core and Copper Loss. A diagram showing the core loss 
and the copper loss in watts for transformers giving outputs from 
1 to 5 kilowatts is given in Fig. 48. By consulting the curve 
showing the iron loss, it will be seen that for a transformer giving 
an output of 2^ kilowatts, the core loss will be about 33 watts. 
The curve for the full-load copper loss reads about 55 watts for 
2| kilowatts output. 

The output is 2500 watts and the total loss is 55+33, or 88 
watts. The input is therefore 2588 and the full-load efficiency is 



V = 



2500 



= 0.966, or 96.6 per cent 



2588 
A transformer having an output of 5 kilowatts has a core loss 

234 



DESIGN OF SMALL TRANSFORMERS 



91 



of 45 watts and a full-load copper loss of 93 watts. The efHciency 
in this case is 

5000 



5138 



= 0.973, or 97.3 per cent 



Fig. 48 may serve as a guide in making assumptions for losses and 
efficiencies in the design and building of transformers for high 
pressures of from 2000 to 3000 volts and ratios of transformation 
of 10 or 20 to 1. 

Cost of Core Loss. The experimenter is of course interested 
in the cost of operating his transformer and the question of the 
cost of the core loss will be considered. In all transformers the 
core loss is very nearly constant at all loads, which means at 



\dO 


-■ 


- — 


— 


- 


n 


— 










— 




— 








- 


- 


- 


n 






— 








- 
















-- 


"T- 


— 








— r- 
























































































































































































































































































































































































































^ > 




















































































































































































































1 




















f 




^ 












































































\ 


0^ 












































































p 


^ 












































































c 










































i 




































. 


^ 














































































^ 


' 












































































^■"^ 






















































40 





__ 


_ 


_ 




_ 










\ \ 




^ 








_ 


_ 


_ 


_ 






_ 








_ 
















_^ 


-A 


_ 












-- 


-- 


- 


- 


-1 


- 










i: 




- 




^_ 




=< 


J 


= 


- 






■^ 








- 
















-- 




- 
























^ 


f^ 






[o 


s 


LC 




































































^ 










I 


T 






























































^ 


? 












































































^? 


f" 


















































































y 


















































































' 












- 










-i 




_ 












_ 


_ 






_ 








_ 










J 






__ 





_ 











Z 3 

.■RATED OUTPUT IN WL0WATT5 



Fig. 48. Copper Losses in Transformers 

one-half the normal rated output as well as at double the normal 
output or at zero useful output. Suppose that the cost of the 
core loss in the 110- volt to 10,000-volt transformer is considered. 
The core loss in this case is 55 watts. If this transformer were 
connected with the supply mains for 1 hour, the total electrical 
energy supplied would be 55 watt hours, or 0.055 kilowatt hour. 
If the rate for electrical energy is, say, 10 cents per kilowatt hour, 
the cost of the core loss during 1 hour will be 0.055X10, or 0.55 
cent, or slightly over | cent. The cost of the core loss for 10 
hours would be 10X0.55, or 5.5 cents. The core loss, at the 
charge rate of 10 cents per kilowatt hour, will always be 0.55 cent 



235 



92 DESIGN OF SMALL TRANSFORMERS 

per hour whether there is a useful output from the transformer 
or the useful output is zero. 

Cost of Copper Loss. For the same transformer the cost of 
the copper loss at a useful output of 1 kilowatt will be the same 
as that of the core loss just figured. At less output, however, the 
copper loss will be less. The core loss in a transformer is a 
constant, while the copper loss is a variable. 

Cost of Operation of Transformer. It will be interesting to 
figure the cost of operating this transformer during 1 hour. The 
useful output is 1000 watts, and since its efficiency is 90 per cent, 
the input is 1111 watts. Since the input to the primary always 
determines the cost of operation at any useful output whatever, if 
this transformer operates during 1 hour at a useful output of 
1000 watts, the cost must depend upon the cost of the input of 
1111 watts for that period of time. The electrical energy input 
during 1 hour will be 1111 watt hours or 1.111 kilowatt hours. 
At a 10-cent rate the cost will be 11.1 cents; that is, a useful 
output of 1 kilowatt hour will cost 11.1 cents. 

The total cost may be divided up as follows: 

Paid for core loss $0.0055 

Paid for copper loss 0055 

* Paid for useful output 10 



Total cost (cents) HI 



Now, in designing this transformer, the efficiency might have 
been assumed as 95 per cent if it had been so desired. The cost 

of operation during 1 hour would then have been XlOX 



1 



0.95 1000* 

or 10.52 cents at a rated output of 1000 watts. However, if the 
efficiency had been assumed to be 95 per cent, the design would 
have been different and the transformer would have cost more; 
it would have had more material in its make-up. 

The apportionment of the loss might also have been made 
differently. Any erratic assumption will always be attended by a 
corresponding result when the design is worked out, and if the 
results are inconsistent or undesirable, then a new design must be 
worked out with a different assumption. 

Flux Density. The assumption as to the flux density is an 
important one, and a good guide might be, never to assume much 



236 



DESIGN OF SMALL TRANSFORMERS 



93 



in excess of 5000 lines per square centimeter when ordinary grades 
of iron are to be used in constructing transformers to be con- 
nected with 60-cycle circuits. 

A diagram illustrating approximately the total magnetic flux 
corresponding with useful outputs in kilowatts is shown in Fig. 49 

0.3 



0.7 



06 



O.S 



0.4 



O.d 



^5 



0.1 











/ 














/ 










/ 


/ 












A 



































































/ Z ^ ^ S 6 

Fig. 49. Magnetic Flux for Transformers Used for Electric Illumination 



for transformers that are to operate electric lamps for purposes of 
illumination. The relations here shown do not apply to trans- 
formers designed for other purposes. 

In Fig. 49 values of useful power output in kilowatts are 
plotted along the horizontal line, while corresponding values of 
total core flux are -plotted vertically in mega maxwells, or in 



237 



94 DESIGN OF SMALL TRANSFORMERS 

millions of maxwells. Since M denotes the flux density per square 
centimeter of core section, the total flux in the core section must 
be the flux per square centimeter multiplied by the number of 
square centimeters in the sectional area of core which is denoted 
by ^. If the Greek letter $ (phi) be used to denote the total 
core flux, then 

The fundamental equation employed in transformer design may 
be expressed in terms of $, as 



iV.= 



lO^^p 



Also since -\/2X7r equals a/2 X3. 1416, or 4.44, the fundamental 
equation may be written 

'^ 4.44/$ 

a form often seen in textbooks. 

Effect of Frequency on Design. It is evident by consulting 
the equation just given that as the frequency of the supply pres- 
sure increases, the necessary number of primary turns for a given 
constant pressure Ep decreases. While a 110- volt transformer 
designed for 60 cycles might safely be used on a 110-volt 100- 
cycle circuit, it could not be safely used on a 110-volt 25-cycle 
circuit. A transformer for 110 volts output, 25 cycles, will 
require more primary turns than will be required on the same core 
for 110 volts output, 60 cycles. 

No=Load Primary Loss. When a transformer has its primary 
connected with a given supply main but no useful output is 
being delivered from the secondary, the primary current is very 
small and is called the exciting current. This small exciting cur- 
rent in the primary simply furnishes the power to balance the 
core loss and whatever very small Rplp^ loss there may be in 
the primary windings. Since the exciting current is very small 
in any size of transformer in comparison with the normal current 
in the primary, the no-load primary wire loss is very small 
indeed compared with the Rplp^ primary loss when the trans- 
former is delivering its rated useful output. In other words, at 



DESIGN OF SMALL TRANSFORMERS 95 

no load the input to the transformer, and therefore the cost, is 
chiefly determined by the core loss. 

Financial Efficiency of Transformer. It is evident that the 
design of transformers depends upon the financial conditions that 
may be assumed quite as much as upon certain electrical condi- 
tions. For a maximum financial return from commercial trans- 
formers in operation, the question of the relation of one loss to 
another needs very careful consideration and calculation. For 
commercial operation the financial efficiency of any of the trans- 
formers discussed in this text may be found when its all-day 
efficiency is figured and when the cost of producing electrical 
energy is known or the charge of operation per kilowatt hour 
is known. 



239 



DESIGN OF ELECTROMAGNETS 
AND INDUCTION COILS 



ELECTROMAGNETS 

Types of Magnets. Magnets are divided into two general types, 
permanent magnets and electromagnets. Permanent magnets are 
those made of hardened steel, which, when once magnetized, 
retains its magnetism for a long time. The commonest forms of 
permanent magnets are the bar and the horseshoe magnets. 

Electroniagnets. Electromagnets are those in which the 
magnetic field is produced by an electric current. Although 
electromagnetic effects can be obtained by simply passing a cur- 
rent through a coil of wire, the wire is usually wound around a 
soft-iron core for the reason that the magnetic effect can thus be 
greatly increased. Soft iron is used for electromagnets because it 
does not retain its magnetism after the current is shut off. It is this 
property of being able to produce magnetic force and have it disappear 
at will which makes the electromagnet so valuable. Take for example 
the electric bell. If the core retained its magnetism, the clapper 
would not be released and the bell would only tap once, instead of 
ringing as long as the button is held down. The telegraph relay 
is another application of the electromagnet. When the operator 
presses his key, the clapper of the relay is attracted to the core, 
and when the key is released, the clapper is also released. With 
a permanent magnet the telegraph would be impossible because 
the clapper would not be released after it was once attracted. 
There are a great many other uses of the electromagnet, some of 
which will be described under the head Electromagnet iVpplica- 
tions. 

THEORY OF MAGNETIC CIRCUIT 

Law of Magnetic Circuit. The operation of electromagnets 
depends on magnetic attraction, and it is necessary to know the 
laws which govern the magnetic circuit before the action of 
electromagnets can be understood. 



241 



ELECTROMAGNETS AND COILS 



The fundamental law of the magnetic circuit is the same in 
form as that of the electric circuit. According to Ohm's law 

E = RI 

for the electric circuit; that is, the electromotive force E required 
to force a current I through an electric circuit is proportional to 
the resistance R. The law governing the magnetic circuit is the 
same in form and can be written 

NI=0lO (1) 

in which NI is the magnetomotive force; O is the flux, or lines 
of force; and 01 is the reluctance of the magnetic circuit. Equa- 
tion (1) simply means that the magnetomotive force NI required 
to force a flux O through a magnetic circuit is proportional to 
the reluctance 01 of the circuit; in other words, the cause is pro- 
portional to the effect. 

Magnetomotive Force. The magnetomotive force (m.m.f.) is 
the force which causes the flux in a magnetic circuit. When cur- 
rent flows through a conductor, a mag- 
netic field is set up about the conductor 
as shown in Fig. L In the illustration 
the current is assumed to be flowing 
toward the paper, and the direction of 
the field is indicated by the arrows. The 
strength of this magnetic field depends 
on the current, that is, 10 amperes will 
produce a field twice as strong as 5 
amperes. Now if the conductor is bent 
to form a circle, the same number of 
lines of force will flow as before, but they will be concentrated inside 
the circle. Fig. 2 shows a cross-section of a turn of wire and also 
shows the concentration of the lines of force at the center. If, 
instead of one turn of wire, two turns of the same diameter as 
before are used, the lines of force will be doubled because the second 
turn of wire will cause just as many lines of force as the first. It 
will therefore be seen that the number of lines of force depends not 
only on the current in a conductor but also on the number of turns. 
A current of 10 amperes flowing through one turn of wire will cause 
the same flux as 5 amperes and two turns of wire. For this 




Fig. 1. Magnetic Field 
about a Wire 



242 



%i 



ELECTROMAGNETS AND COILS 



ason the magnetomotive force is expressed in ampere turns, the 
product of the current times the turns. 

Flux. The flux in a magnetic circuit is the total number of lines 
of force passing through any complete crQss-section perpendicular to 
the direction of the lines of force. It should be borne in mind that 
the total flux through any complete cross-section is the same. The 
lines of force do not go wandering off into space, but each line 
forms a complete linkage with the coil producing it. 

The unit of flux is the line of force, or the maxwell, as it is 
generally called. When a conductor moves across a uniform 
magnetic field at a uni- 
form rate, an e.m.f. of 1 
volt is induced if the 
conductor cuts across 
100,000,000 maxwells per 
second. 

Reluctance. Reluc- 
tance in the magnetic 
circuit corresponds to 
resistance in the electric 
circuit; it is a measure 
of the opposition which 
must be overcome in 
forcing a flux through 
any magnetic path. The reluctance of air and all other nonmag- 
netic materials ip the same. The reluctance of iron, however, is very 
low, and therefore iron is used in nearly all magnetic structures. 
The unit of reluctance is the rel, which is defined as the 
reluctance of a magnetic circuit through which a magnetomotive 
force of 1 ampere turn will force a flux of 1 maxwell. For air 
and other nonmagnetic materials, the rel is the reluctance of a 
magnetic circuit 1 square inch in cross-section and 3.19 inches long. 
From equation (I) or from the above definition 

Ampere turns = rels X maxwells 

Equation (1) can also be written as 

NI 




Fig. 2. Magnetic Field about a Turn of Wire 



(R = 



O 



(2) 



243 



4 ELECTROMAGNETS AND COILS 

or 

* The reluctance of a magnetic path is proportional to its 
length; that is, the reluctance of a 1-inch rod 2 inches long will 
be one-half that of a 1-inch rod 4 inches long. The reluctance 
is inversely proportional to the cross-sectional area of the magnetic 
path — the greater the area, the less the reluctance. The formula 
for reluctance is 

01=—, (4) 

where / is the length of the magnetic path in inches, A is the 
cross-sectional area in square inches, and /x is the permeability of 
the material. 

Permeability. As previously stated, for air the rel is the 
reluctance of a magnetic circuit 1 square inch in cross-section and 
3.19 inches long, and if these values are substituted in equation 
(4) , it will be found that the permeability of non-magnetic naate- . 
rials is 3.19. The permeability of iron is much higher and depends 
on the quality of iron and also on how much flux is flowing 
through the iron. 

Exam-pie 1. A magnetic circuit has a reluctance of 3 rels. How many 
ampere turns (a.t.) will be required to produce a flux of 500 maxwells? 
From equation (1) since (>? is 3 and ^ is 500, therefore 

i\^7 = 3X500 = 1500 a.t. 

Ans. 1500 a.t. 
Example 2. How many Unes of force will be produced in a magnetic 
circuit with a reluctance of 5 rels if the magnetomotive force is 1650 ampere 
turns? 

Ans. 330 maxwells 
Example 3. A magnetomotive force of 2300 ampere turns produces a 
flux of 1500 maxwells. What is the reluctance of the circuit? 

An%. 1.53 rels 
Example J/-. What is the reluctance of a block of wood 8 inches long, 
4 inches wide, and 2 inches high? 

Equation (4) appUes and I equals 8 inches, /i (wood being nonmagnetic) 
equals 3.19, and A equals 2X4, or 8 square inches. 
Therefore 



^=3-lf^8=^-^l^^^ 



244 



Ans. 0.31 rel 



ELECTROMAGNETS AND COILS 5 

Example 5. A wooden rod has a cross-sectional area of 2 square inches 
and its reluctance is 1 rel. How long is it? 

Arts. 6.38 in. 

Reluctance of Magnetic Paths in Series. In a great many 
magnetic circuits the magnetic path is not uniform. The area may 
vary in different parts of the circuit and different parts of the 
circuit may be made up of different materials. In cases of this 
kind the circuit can be broken up, the reluctances of the different 
parts calculated from equation (4), and the results then added 
together to get the total reluctance. 

Reluctance of Magnetic Paths in Parallel. Reluctance is a 
measure of the difficulty of forcing flux through a given path, 
while permeance is a measure of the ease with which it can be 
forced through. Permeance (P is therefore the reciprocal of reluc- 
tance 01 and, therefore^ 

When a magnetic circuit has several paths in parallel, the perme- 
ances of the several paths are added together to get the total 
permeance of the circuit. After the total permeance has been 
found, the total reluctance can be found from 

^ = ~ . (6) 

From equation (5) it will be seen that the permeability ^ of a 
material is the permeance of a unit cube of the material. 

Example 6. A magnetic circuit has two paths in parallel. The first 
is 8 inches long and 2 square inches in cross-section; the second is 6 inches long 
and 1.5 square inches in cross-section; fx in both cases is 4000. What is the 
total reluctance of the path? 

Substituting the values given in equation (5), the permeance of the first 
circuit is 



and that of the second circuit 



^ 4000X2 ^^^^ 
6P= — = 1000 



_ 4000 XU ,^^^ 

(P = = 1000 

6 



The total permeance is then 1000 + 1000, or 2000. 
and from equation (6) the reluctance is 



(R = =0.0005 rel 

2000 Ans. 0.0005 rel 



245 



6 ELECTROMAGNETS AND COILS 

By substituting in equation (1) the value of (R in equation 
(4) the following equation is obtained: 

mJ^^ (7) 

This equation may also be written 

Example 7. How many ampere turns will be required to force a flux 
of 5000 lines through the rod in example 5? 
Substituting values in equation (7) gives 

,^^ 6.38X5000 ^^^^ 

NI = — - =5000 a.t. 

3.19X2 

Ans. 5000 a.t. 
Magnetic Intensity. Of course, equation (7) only holds true 
when the magnetic path is uniform; if the path is irregular, it is 
necessary to divide it into parts which are uniform and compute 
the ampere turns for each part. In making calculations of this 
kind it is convenient to use the ampere turns consumed per inch 
length of magnetic path, or the magnetic intensity, as it is called. 
Magnetic intensity is designated by 9^, and as the path over 
which the intensity is taken is assumed to be uniform 

NI 

^=^ . (9) 

Flux Density. The flux density is the lines of force per square 
inch cross-section and is designated by /3. The flux density in a 
uniform field is of course equal to the total flux divided by the 
cross-sectional area or, expressed in symbols 

/§ = ^ (10) 

Relations of B and H. If yS and ^ are substituted in equation 
(8) for their respective values in equations (9) and (10), the fol- 
lowing equation results: 

/3 = fx^ (11) 

Equation (11) may also be written 

M = | (12) 



246 



ELECTROMAGNETS AND COILS 



or 



^ = 






(13) 



H for air and nonmagnetic materials is 3.19, therefore for air the 
flux density /§ is always 3.19 times the ampere turns per inch d^. 
Simple Magnetic Circuit. The simplest form of magnetic 
circuit is an iron ring on which has been wound a coil of wire in 
the form of a helix, Fig. 3. There is practically no leakage and 
the cross-sectional area is constant, and therefore equations (U) 
(12) and (13) can be used directly for this type of circuit. 

Example 8. Assume the mean diameter D of the coil in Fig. 3 to be 
6 inches and the cross-sectional area of the magnetic path 1 square inch. How 
many ampere turns will be required to produce a flux of 500 maxwells, assum- 
ing that the coil has an air core? 

As the diameter of the coil is 6 inches, 
the total length I of the magnetic path is 
6X3.1416, or 18.85 inches; the value of * is 
500; A is 1; and /z for air is 3.19. Therefore 
by substituting values equation (7) becomes 



A^/ = 



18.85X500 
3.19X1 



= 2952 a.t. 
Ans. 2952 a.t. 




Fig. 3. Coil of Wire 
around a Ring 



Example 9. It is desired to produce a 

flux of 200 maxwells in the coil described in 

example 8, and the current is 10 amperes. 

How many turns should be wound on the coil? 

Ans. 118 turns 

Example 10. Assume the diameter D of 
the coil in Fig. 3 to be 5 inches and the cross- 
sectional area of the path to be 0.785 square 
inch. If the coil is wound with 90 turns and the current is 6 amperes, what flux 
will be produced? 

Ans. 86.2 maxwells 

Example 11. A coil with the same dimensions as the one in example 10 
is wound with 600 turns. The resistance of the coil is 0.9 ohm. What flux 
will be produced if an e.m.f. of 4.5 volts is applied at the terminals of the coil? 

Ans. 479 maxwells 

A straight helix is shown in Fig. 4, and the dotted lines 
indicate the directions taken by the lines of force. With this type 
of coil all the lines of force do not link with all the turns as was 
the case in Fig. 3. With an equal number of ampere turns the 
flux produced in the middle of the coil in Fig. 4 is the same as 



247 



8 ELECTROMAGNETS AND COILS 

the flux in Fig. 3, but the flux at the ends of the coil in Fig. 4 is 
only one-half that at the center. 




Fig. 4. Magnetic Field about a Coil 











— 
























- 




1 


_ 


.^ 


— 


















■ 


















nT 


IS 


Hi 




— 




-" 






































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^i. 


IP* 


















































1 


r> 


1^ 






















































/ 
























*-- 


e 


• /. 



























^ 




^ 


















/^ 


ifi 


r 


_^ 


7^ 








'^ 
























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f 
















-^ 


isi. 


■^ 








































/ 










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^ 


' 














































/ 








^ 


























































/ 




















































\ 






/ 






















































^ 6o ooo • 




i 


r~ 






















































Vl 




/ 
























































^f 




























































! 
























rr 


7 


IT 


?^ 









— 


-" 


"^ 


















^ 




















^ 


4i 




















































^ 


^ 












































n^ 




























































•^ 






^ 


























































/^ 














































































































































_ 


L 




_ 
















L_ 













Fig. 5. Magnetization Curves 



Effect of Iron in the Magnetic Circuit. The permeability of 
nearly all substances is the same as that of air at normal atmos- 
pheric temperature and pressure, but nickel and chromium are 
slightly magnetic, and the permeability of iron is a great many 



248 



ELECTROMAGNETS AND COILS 9 

times as large as that of any other known material. For this 
reason the effect of iron in the magnetic circuit is to greatly 
increase the flux. Fig. 5 shows the magnetization curves of cast 
iron and of cast steel. It will be seen from the curves that for 
the smaller values of ^, the flux density M is very nearly pro- 
portional to ^. As the magnetomotive force ^ is increased, how- 
ever, M does not increase as fast as d^ because the iron becomes 

saturated. From this it will be seen that for iron — is not con- 

stant; that is, the permeability of iron is a variable quantity, 
depending not only on the quality of the iron but also on the point 
on the magnetization curve at which it is being worked. 

Example 12. When M equals 18,000 maxwells per square inch, what is 
the value of ix for the sample of cast iron the curve of which is shown in Fig. 5? 

From the curve it is found that for cast iron, when M is 18,000, ^ equals 
15. Substituting these values in equation (12) gives 

18000 __ 

/i= = 1200 

15 

Ans. 1200 

Example 13. When the ampere turns per inch are 60, what is the value 

of n for the sample of cast steel the curve of which is shown in Fig. 5? 

Ans. 1600 

Example H-. If the coil in example 10 had a cast-iron core instead of 

an air core, how many ampere turns would be required to produce a flux of 

24,000 maxwells? 

$ 24000 
yS = — - =-— — -= 30573 maxwells 
A 0.785 

From the cast-iron curve in Fig. 5 when M equals 30573, ^ equals 36. From 

equation (9) NI = m 

Therefore AT/ = 36X5X3.14 = 565 a.t. 

Ans. 565 a.t. 

COIL CONSTRUCTION 
General Considerations. In designing a coil for an electro- 
magnet there are a number of things which must be considered. 
The ampere turns required by the solenoid are usually known, and 
the first requirement is to produce a coil which will have the cor- 
rect number of ampere turns. The coil must also be designed so 
that it will not overheat and the heating depends on the duty; 
that is, a coil will get hotter when it is in circuit all the time than 
when it is in circuit only part of the time. For this reason a 



249 




10 ELECTROMAGNETS AND COILS 

larger coil is required for continuous service than for part-time 
service, or intermittent duty, as it is usually called. 

Kinds of Wire. The kind of wire to be used must also be 
determined. The kinds of magnet wire in general use at the 
present time are, single cotton-covered, double cotton-covered, 
single silk-covered, enamel, and asbestos-covered. Single cotton- 
covered wire is the cheapest and is the kind most generally used. 
The thickness of the insulation varies from 0.004 to 0.006 inch 
depending on the size of the wire, being thinner for wire of small 
diameter. 

Double cotton-covered wire is used principally on coils in 
which the voltage between layers is high and on alternating cur- 
rent. The heavier insulation is valuable on 
alternating current because there is usually some 
vibration, which might cause turns to short- 
circuit and burn out the coil. When turns are 
shorted on an a.c. coil, the effect is the same as 
on a short-circuited secondary on a transformer 
Fig. 6. Cross-Section and a burncd-out coil usually results. The 

of Winding , . , "^ 

thickness of insulation on double cotton-covered 
(d.c.c.) wire varies from 0.008 to 0.012, depending on the size of 
the wire and is just twice that on single cotton-covered (s.c.c.) wire. 
Therefore a coil wound with d.c.c. wire is larger than one of the 
same number of turns wound with s.c.c. wire. This can easily be 
seen from Fig. 6, which shows the cross-section of a winding. Sup- 
pose that this coil is wound with 100 turns of No. 20 s.c.c. wire. If 
this winding should be replaced by No. 20 d.c.c. wire, it would, of 
course, be impossible to get 100 turns in the same space because 
the diameter of the wire is larger; suppose that 80 turns of d.c.c. 
wire could be wound in the space. The resistance of the new coil 
w^ould then be only 0.8 as large as that of the first and the coil 
would overheat because 

^-,. , . volts2 

V V atts = — -. 

resistance 

The watts consumed in the second coil would be approximately 
25 per cent higher than those consumed in the first, and in order 
to keep the coil from overheating, it would be necessary to make 
it larger. 



250 



4 



I 



ELECTROMAGNETS AND COILS 



11 



Silk-covered wire is more expensive than cotton but is valua- 
ble because silk is a better insulator and because the silk insula- 
tion is thinner and therefore gives a more efficient winding than 
cotton. 

Asbestos-covered wire is used for high-temperature coils, in 
which it has been found that cotton insulation chars. The thick- 
ness of asbestos insulation is usually about the same as that of 
d.c.c. wire. 

In recent years enamel w4re has replaced the other types to 
a large extent. Enamel wire will stand higher temperatures than 
cotton or silk, and its insulation is also thinner than that of any 
other type and more efficient windings can therefore be produced. 

Cost. The cost should be 
considered in designing an electro- 
magnet. If two designs will do 
a given job equally well, the one 
which costs the least should, of 
course, be chosen. The weight 
of copper in a coil should be 
checked closely, because the cost 
of copper is a large part of the 
cost of an electromagnet. 

Bobbin=Wound Coils. In 
making bobbin-wound coils the 
bobbin is first built up and the 
wire is then wound on it. A great 
many different types of bobbins 
have been made, one of which is 
shown in Fig. 7. ^ is a copper tube spun over at the ends and 
B and C are the coil heads, which are made of fiber or other 
insulating material. The head C is provided with a slot for bring- 
ing out the inside lead, and Z) is a washer used to insulate the 
inside lead from the rest of the winding. E is an insulating 
material such as empire cloth, which is wound around the tube A 
to insulate the winding from the tube. After the wire is wound on 
the bobbin, the terminals are soldered to the two ends and the 
winding is protected by winding tape or fish cord around it. The 
outside is then painted with insulating varnish. 




Fig. 7. Bobbin-Wound Coil 



2S1 



12 



ELECTROMAGNETS AND COILS 



Another very simple type of bobbin can be made by using a 
tube of insulating material instead of the copper tube A. The 
tube is threaded at both ends, and the coil heads are also threaded 
and the threads filled with insulating varnish. 

Some manufacturers are now making bobbins of molded 
insulation, and very satisfactory results are obtained with this con- 
struction. 

Form=Wound Coils. During the last few years the bobbin- 
wound coil has been used very little and is being replaced by the 
form-w^ound coil, which, as the name implies, is wound on a form, 
usually made of steel. In building a form- wound coil the form is 
put in the lathe and the core wrapped with insulating material, 

usually insulating cloth or heavy 
paper. The required number of 
turns of wire are then wound on, 
and the coil is removed from the 
form and taped and the terminals 
soldered and fastened. Insulating 
washers are placed on the ends of 
the coil above and below the inside 
lead. After the winding is finished, 
these coils are usually treated to 
improve the insulation. Sometimes 
they are dipped in insulating var- 
nish and baked, but generally they are impregnated by the vacuum 
process, which not only improves the insulation but helps to con- 
duct away the heat from the inside. 

A cross-section of a typical form-wound coil is shown in Fig. 8. 
A great many coils are now being made by automatic machin- 
ery, and they are a great deal better than any which can be made 
by hand as there is very little chance, with this method, of get- 
ting a high voltage between turns. When coils are made by hand 
it is hard to get an even winding, and it is usually necessary to 
put a thin layer of paper between the layers of wire to make sure 
that the winding will not break down between turns. On long 
thin coils on which the voltage between turns is Hable to be very 
high, the coil can be made in two or more sections and the sec- 
tions connected in .series. 





1 




1 

9 


o 
o 



Fig. 



Form-Wound Coil 



252 



ELECTROMAGNETS AND COILS 



13 



B WINDING CALCULATION 

^^ Cylindrical Coils. In practice nearly all coils are of the cylin- 
drical type as shown in Fig. 9. In this illustration L is the length 
of the coil, M is the average diameter of the coil, T is the thick- 
ness of the winding, and Pa is the average length of all the turns. 
The cross-sectional area of the winding will then be TxL, and 
the volume of the winding V will be equal to this cross-sectional 
area times the average length of turns, which expressed in symbols is 

V=TxLxPa (14) 

The number of turns of wire on the winding N will be equal to 
the cross-sectional area times the number of 
turns per square inch Na, or 

N=TxLxNa (15) 

The value of Na varies for different sizes 
of wire and can be obtained from Tables I and II. 

The total length of wire in a winding can 
be found by multiplying the total number of 
turns by the average length of turn. Then if 
the resistance per inch of length is known the 
total resistance of the coil R can be calculated 
because 




Fig. 9. Cylindrical Coil 



R=NP,Ri 



(16) 



where Ri is the resistance per inch of wire. The values of Ri 
for different sizes of wire are given in Tables I and II. The total 
weight of wire in a coil can be found by dividing the total resist- 
ance of the coil by the ohms per pound of the wire used: 



Weight of wire = — lb. 



(17) 



where R is the total resistance of the winding and R^^ is the ohms 
per pound. The values of R^ for the different sizes of wire are 
given in Tables I and II. Another way to find the weight of wire 
is to multiply the volume by the pounds per cubic inch. 

Example 15. A coil is 4 inches long and the depth of the winding is 
1 inch. How many turns will the winding have if wound with No. 26 s.c.c. 
wire? 

From Table I it will be found that for No. 26 s.c.c. wire the turns per 
square inch of winding cross-section are 2385. 



253 



I 



14 




ELECTROMAGNETS AND COILS 












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ir^ Oi CO 


CO O Oi 


l-H rH CO 


CO CO 1-H 


-rtl OO CO 


CO CO I-H 


CO CO (M 


CO Tfi (M 










^-^(M 


CO '^ '^ 


COI>00 


OCO CO 


CiCOOO 


^t-t- 


tO^CJ 


Ci(Mt^ 














l-H l-H I-H 


rH(M (N 


CO CO '^ 


toco CO 


1> 05 05 




qiO'* 


PPP 


oo^ 


OOO 


to to p 


PPP 


OOO 


PP9 


ppp 


ppp 


ppp 


lo locd 


i>I ood 


d^oj 


•^ tot-l 


d i-^ -ri^ 


t^ d r-l 


tod'* 


t^ CO 00 


'^ 1>I to 


oi t6c<i 


ddd 








T-H tH rH 


i-H (M C<) 


(M(M CO 


CO '^ -* 


Tt^ tOiO 


C0C0 1> 


00 GO Oi 


0500 


^ ^ 






















T— 1 l-H 




iOO<M 


CO CO <M 


iOOO o 


O CO Tft 


00 -^ •<# 


OOOiCO 


OiOO 


OOO 


OCSI to 


OO^ 


^^^ 


Sfeo 


<M as Tfi 


OCiCs 


t-'^Oi 


CO O T-H 


OS Oi to 


oqi>.a5 


'^Ot^ 


^_qoq 


r-JrHCX) 


iO(M (M 


r^f^^ 


rl?'^ i6 


cdcdt-^ 


oodd 


csicoio 


dood 


c<i TjH d 


d^d 


T-H lood 


■TtH d (m" 


t^ (M* to 


00 00 00 


H ^ 






1-1 




T-H r-H (M 


(M (M (M 


CO CO CO 


TjH TjH 'stH 


to to CO 


COt-t- 


S^l 
^«^ 


"^ lOCD 


t^00O5 


Ot-hC^ 


CO-* to 


COl>00 


GiOi-i 


(MCO'* 


to CO t^ 


00 050 


^(MCO 


Tt^lOCO 






r-t y—i i—{ 




1—1 I-H I-H 


i-i(M(M 


(M(N<N 


(M(N<M 


(M(NCO 


CO CO CO 


CO CO CO 



254 



ELECTROMAGNETS AND COILS 



15 



Ohms 
Per 
Inch 

(R/) 


0000535 
0000679 
0000828 


i-H i-H i-H <N (M CO 


000420 
000530 
000670 


000845 
001065 
001343 


1— 1 (M (N CO -^ lO 


00680 
00856 
01082 


01368 

0172 

0217 


0274 
0345 
0435 


^CN(N 
lOOt^ 




ooo 


ooo 


ooo 


ooo 


ooo 


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ooo 


ooo 


ooo 


ooo 


l-cs 


ooo 


»oo^ 


ooo 


OOO 


lO >0 lO 


lOiOiO 


lOiOO 


OiOO 


OOiO 


ooo 


lOiLOO 


Ft 


OiOi OO 

CO coco 


s§s 


OiOiO 
COCOI> 


0005 
!>!>• CO 


oot-t^ 

CO CO CO 


coco CO 

CO (£)(© 


ggg 


s^s 


eo^(N 
coco CO 


coSS 


sgg 


CD T3^ 


1-H ^ (N 

ooo 


gis8 

OO-H 


lOi-io 

1-1 CO »o 


g^§ 


<M 1-1 tH 


fegg 


OCO(N 


ooo 


OOO 


ooo 


ooo 


ooo 


ooo 


OOO 


Oi-KN 


coiooo 


(MO(M 

i-i(MCO 


(MC^i-i 

lOOOCO 

1—1 


Oi (M CO 
OCO(M 
(MCOiO 


cO(N O 

COCO (N 

00 CO 1-1 

1-1 (N 


CO lOOO 


ll§ 

■<* ■* 1— ( 
CO 1-* "* 
i-i(NCO 


-g 


























CO-* CO 
lO -^ CO 
(M (M (N 


CO 00 CO 

CO CO CO 

01 (M (M 


1-H (M CO 
COCO(M 
Cfl <M (M 


(N (M (M 




(M C^ <M 


CO to lO 


CO (NO 
(N (N (N 


1>OLO 


lO to CO 


8SS 

(N (N 1-1 




OOO 


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OOO 


odd 


OOO 


OOO 


OOO 


ooo 


ooo 


OOO 


j3 


III 


ooo 


""^ t^ 1-1 

OO-^ 


(M CiCO 


O'^O 
CM CO O 
1> 1-1 00 


<MO5 00 
OO^O 


^^^ 


COi-irfH 


ooo 


ooo 


ooo 




OOO 


OOO 


OOO 


OOO 


Oi-*i-H 


<NtJH t^ 


;:^^85 


^ OO 
"* t^ 1-1 


COOiCO 

t^t^ CO 

1-1 (N "* 


•<*TjHCO 

00 Oi (N 
COOI-- 

T-l 1—1 


CO(N O 
<3iCOt^ 
CO CO 1:^ 

(N TtH CO 


J3 


ot^co 


ooo 


OOO 


OOO 


ooo 


OOO 


OOO 


OOO 


ooo 


ooo 


OOO 


Turns 
Per 
Square I 
(Na) 


CO CO CO 
lOCOOO 


COLO 00 
O CO CO 
1— 1 I— ( r-l 


rH t>- Ttl 

(Mc^eo 


CO lOi-l 
CO -^ 00 
^ lOcO 


oooco 
1—1 1—1 


S8^ 


OiOO 
(M ^ lO 

CO rH (M 

CO'^iO 


OiOO 

1-1 1^ o 
CO ooo 

1—1 


g8S 

CO (N Oi 
(NcO(3l 


(NCOCO 


0:1 (N t^ 


£ ^ 


OOt^ 00 


ocooo 


t-tOl> 


O00>O 


i>ioco 


o»oo 


oo»o 


ooo 


ooo 


ooo 


000 




t^ 00 Oi 


i-H C? CO 

T— 1 T— 1 1— 1 


lOt-Oi 

1— * T-H T— 1 


(M ■* b- 


CO CO CO 


COOO"* 
TJ^Tt^ lO 


cocot^ 


^^1 


Ci'^Oi 

i-ieoTt< 

T-( tH ,-( 


1-1 1-1 (N 


CO >o 10 

CO COOS 
(N(N(N 




OOCO ^ 


COOCN 


i>io»o 


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t-ooo 


1> CO CO 


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lOOO 


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000 


COI>00 


05 i-KM 
1— < tH 


CO iOI> 
1—1 1—1 rH 


i-i(N<N 


l> rH Ttl 

(NCOCO 


00 CO 00 
CO'*'"* 


Tf ^ 00 

iccoco 


CO CO >o 
l>00Oi 


co^^ 
0(NCO 

1— ( 1— 1 T— 1 


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lOCOOO 

T— 1 1— 1 1— 1 


>-0 CO 
1-^ CO CO 
(N(N(N 




00 050 


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00 05Q 
COCO^ 











255 



16 ELECTROMAGNETS AND COILS 

Substituting in equation (14) gives 

iV = 1 X 4 X 2385 = 9540 turns 

Ans. 9540 turns 
Example 16. What will the resistance of the coil in example 15 be if 
the average length of turn is 6 inches? 

Ans. 194.6 ohms 
Example 17. Find the weight of copper in the coil in example 16. 

Ans. 4 lb. 
Example 18. How many ampere turns will the coil in example 16 have 
if 100 volts is appHed at its terminals? 

Ans. 4902 a.t. 

Procedure in Designing Coil. In designing new coils the 
ampere turns required by the magnetic circuit are usually known 
and the problem is to select the proper size of wire and find the 
proper dimensions for the coil. According to Ohm's law the cur- 
rent which will flow in a coil is equal to 

'=1 

Furthermore ampere turns are the product of amperes times turns. 

E 
Therefore substituting -^ for I 

NE 
NI=^ (18) 

If the value of R found in equation (16) is substituted in equation 
(18), the following is obtained: * 

NE _E_ 

Now if the voltage of the circuit and the ampere turns required 
are known, the length of the average turn and therefore the 
diameter of the coil will depend on Ri, which in turn depends on 
the wire size. 

Size of Wire. It is possible to get the required ampere turns 
with any size of wire, but a coil wound with No. 20, for example, 
will be much smaller than one wound with No. 16. On account 
of the cost, the diameter of the coil should be as small as possible, 
but it must be made large enough so that it does not overheat. 

The temperature of a coil depends on three things: the watts 
due to the resistance of the coil; the length of time the coil is 
in circuit; and the radiating surface of the coil. The temperature 



256 



ELECTROMAGNETS AND COILS 17 

rise usually allowed on coils is 50° C. and it has been found by 
experiment that if a coil for continuous duty is designed to dis- 
sipate 0.7 watt per square inch of cylindrical surface it will not 
overheat. This means that a coil which must dissipate 14 watts 

14 
should have a cylindrical surface of -— — or 20 square inches. 

For half-time or intermittent duty a coil will dissipate 1.4 watts 

per square inch with the same temperature rise. 

The watts consumed W by any d.c. coil are 

W=EI 

which becomes 

E^ E . . 

W=-^ when — - is substituted for I. 

The watts which a coil can be allowed to dissipate continuously 
Wa and still not overheat are 

Wa = {M-\-T)TrXLxO,7 = (M+T)xLx2.2 (20) 

The most efficient coil will be one in which 

W=Wa 

which may be written 

^={M-{-T)XLx2.2 (21) 

Equation (21) can now be used with equation (19) to find the 
correct size of wire. 

Design Suggestions. The following suggestions will be found 
of value in designing coils: 

The ampere turns can be increased by: (1) decreasing the diameter; (2) 
using larger wire. 

The ampere turns can be decreased by: (1) increasing the diameter; (2) 
using smaller wire. 

The watts can be increased by: (1) decreasing the diameter; (2) using 
larger wire; (3) making the coil shorter. 

The watts can be decreased by: (1) increasing the diameter; (2) using 
smaller wire; (3) making the coil larger. 

Example 19. Design a coil for continuous operation on 230 volts direct 
current; 4000 ampere turns are required, the tube over which the coil is to fit 
is If inches in diameter, and the over-all length of the coil is 6^ inches. 

E 230 

From equation (19) PaRi= — = = 0.0575 ohm 

NI 4000 

The diameter of the tube over which the coil is to fit is If inches and, allowing 
for ^-inch insulation around the tube, the inside diameter of the winding will 
be Ixf inches. The average length of turn Pa will then of course be some- 



257 



18 ELECTROMAGNETS AND COILS 

thing greater than TrXlifj or 6.1 inches approximately. If Pa were only 6.1, 

0575 

Ri would be — , or 0.0095. However, Pa is greater than 6.1 and there- 

6.1 

fore Ri is less than 0.0095. Table I shows that Ri for No. 30 s.c.c. wire is 

0.00876 ohm, and for No. 29 s.c.c. wire is 0.00694 ohm. Assume that the coil 

is wound with No. 30 s.c.c. wire. Therefore 

PaRi 0.0575 

P„ = = = 6.6 m. 

'^ Ri 0.00876 

X 3.1416 
r=ikf-inside diameter = 2.1-1.94=0.16 in. 
L=6|— 1^ (insulation) =5.94 in. 

Since from Table I the turns per square inch Na for No. 30 s.c.c. wire are 4713 

i\r = LxTX iVa = 5.94X0.16X4713 = 4480 turns 

72 = PaXi^iXiV=0.0575X4480 = 258 ohms 

„, E^ 230X230 

W = —- = = 205 watts 

R 258 

Tra = (M+r)XLX2.2 = 2.26X5.94X2.2 = 29.5 watts 

This means that the coil wound with No. 30 s.c.c. wire is not large enough in 

diameter to radiate the energy, and therefore larger wire must be used in order 

to increase the size of the coil. 

Assume now that the coil is wound with No. 29 s.c.c. wire. 

^ 0.0575 „ „ . 

Pa= = 8.3 m. 

'^ 0.00694 

M = -''-=-^^ = 2.64in. 
TT 3.1416 

r=ikf -inside diameter = 2.64 -1.94 =0.70 m, 

L = 5.94 in. 

Since from Table I the turns per square inch Na for No. 29 s.c.c. wire are 3760 

i\r = LxTXA^« = 5.94X0.7X3760 = 15634 turns 

i2 = P«Xie/XA^ = 0.0575X15634 = 900 ohms 

_ E^ 230X230 

W=—= = 59 watts 

R 900 

Tr« = (M + r)XLX2.2 = 3.34X5.94X2.2 = 43.5 watts 
It is therefore necessary to still further increase the size of the coil until the 
watts Wa which the coil can safely dissipate are equal to the watts W con- 
sumed by the coil. 

Assume now that the coil is wound with No. 29 enamel wire, which from 
Table II has an Ri of 0.0068 ohm 
„ 0.0575 

M=i^=-M^ = 2.68in. 
TT 3.1416 

r = M -inside diameter = 2.68 -1.94 =0.74 in. 

L = 5.94 in. 



258 



ELECTROMAGNETS AND COILS 19 

Since from Table II the turns per square inch A^^^ for No. 29 enamel wire are 

6510 A^ = Lxrx Ar„ = 5.94X0.74X6510 = 28500 turns 

R = PaRi X N =0.0575 X 28500 = 1640 ohms 

_ ^2 230X230 ^^„ 
TF = -— = ,„,^ — = 32.2 watts 
R 1640 

Tr« = (M + r)XLX2.2 = 3.42X5.94X2.2 = 44.5 watts 

Wa is now greater than W, and therefore the coil wound with 28,500 turns of 

No. 29 enamel wire will give the required ampere turns and will not overheat. 

It will be noticed however that Wa is 44.5 watts and W is only 32.2 watts. 

The coil is therefore actually larger than it needs to be. By decreasing the 

diameter of the coil the radiating surface is decreased and Wa is decreased; 

at the same time W is increased because the resistance of the coil becomes less. 

It is therefore apparent that there wiU be one diameter at which W will equal 

Wa- An equation can be developed for finding this diameter but it is very 

complicated and it is easier to find the correct diameter by trial. It should 

be remembered in choosing the trial values that W will increase faster than 

Wa will decrease. 

The difference between the last-found values of Wa and W is 12.3 watts. 

Assume that about 30 per cent of this difference, or 3.5 watts, wiU be taken 

up due to decrease in Wa 

Tra = 44.5-3.5 = 41 watts = (ikf+T)X 5.94X2.2 

M+T= = 3. 14 in. 

5.94X2.2 

The inside diameter of the coil is still 1.94 inches, therefore 

3.14-1.94 1.20 ^^. 
r= = =0.6 in. 

2 2 

M = 2.54 in. 

p^ = ^M = 3.14X2.54 = 8 in. 

JV = LXTxAr„ = 5.94X0.6X6510 = 23200 turns 

i2 = P«Xi2zXiV = 8X0.0068X23200 = 1260 ohms 

E^ 230X230 

W = — = = 42 watts 

R 1260 

For the last coil Wa is 41 watts and W is 42 watts, which means that the coil 

is not quite large enough. 

Assume that 30 per cent of the difference will be taken up by increase 

in Wa. T7a = 41.3 watts = (M + r)X 5.94X2.2 

M + T = — ^^4— -=3.16 in. 
5.94X2.2 

„ 3.16-1.94 1.22 ^^^ . 

T= = =0.61 m. 

2 2 

M = 2.55 in. 

P« = 3. 14X2.55 = 8.01 in. 

AT = 5.94X0.61X6510 = 23550 turns 

22 = 8.01X0.0068X23550 = 1283 ohms 

2302 

W = =41.2 watts 

1283 

The correct coil is therefore 23,550 turns of No. 29 enamel wire. 

259 



20 



ELECTROMAGNETS AND COILS 



Example 20. What size of enamel wire should be used if a coil of the' 
dimensions and ampere turns of that in example 18 is used on 115 volts? 

Ans. No. 26 enamel wire 

Protected Coils. For some applications it is possible to use 
an intermittent-duty coil for continuous service by inserting 
resistance in series with the coil at the end of the travel of the 
magnet. This is usually accomplished by having a small switch 
on the magnet which normally short-circuits the resistance. At 
the end of the travel of the magnet this switch is opened and the 
resistance inserted. The advantage of this arrangement is that 
for a given number of ampere turns it is possible to use a much 
smaller and therefore less expensive coil. 

Allowance for Heating. The resistance values given in 
Tables I and II are based on copper at 20° C. After a coil has 

been in circuit, however, the temper- 
ature rises, and as stated before, the 
temperature rise allowed is usually 
50° C. When the temperature of cop- 
, per is increased, the resistance is in- 

'*\-b'^ ^.y^^^ creased, and in going from 20° C, to 

70° C, the resistance increases about 20 
per cent. When the resistance of a 
coil is increased, the current, and there- 
fore the ampere turns, are decreased. 
For this reason the ampere turns of a coil to be used on direct 
current should be increased 20 per cent above the amount required 
by the magnetic circuit in order to ensure proper operation after 
the coil heats. 

Square=Core Coils. On a.c. circuits the magnetic structure 
is usually made up of laminated iron. These iron laminations 
must be fastened together rigidly and therefore it is not possible 
to make the cores circular on account of mechanical difficulties. 
The cores of a.c. magnets are oblong or square, Fig. 10. 
For this type of winding the average length of turn is 

P«=2(a+6) + 7rr (22) 

where a and h represent the core dimensions. For a square core 
a equals h and 

P„ = 4a+7rr 




Fig. 10. Square Coil 



260 



I 



ELECTROMAGNETS AND COILS 21 



rhe turns, resistance, and weight of this type of winding can be 
calculated from equations (15), (16), and (17). The equation for 
illowable continuous watts is different, however, and for oblong 
cores is 

Wa = (a-\-b-\-TrT)lAL (23) 

jFor square cores a equals b and 

Wa={2a-{-TrT)lAL 

TYPES OF ELECTROMAGNETS 
Effect of Work to Be Done. There are a great many factors 
to be considered in selecting an electromagnet for any given appli- 
cation. The work to be done is the most important factor. It 
will, of course, require a larger magnet to do 20 inch pounds of 
work than it will to do 10 inch pounds of work. Even when the 
total amount of work to be done on two different jobs is the same, 
the design of the electromagnets may be altogether different. For 
instance, take the case of two magnets, one of which is to lift 
5 pounds a distance of 2 inches and the other is to lift 2 pounds 
through 5 inches; the first magnet would be comparatively short 
and thick, while the second magnet would be long and thin. The 
distribution of the load will also affect the design of an electro- 
magnet. On some magnet applications the load at the start of 
the stroke is very light and increases as the stroke progresses; this 
condition is quite different from that when the load is the same 
throughout the stroke, and it requires a different type of magnet. 
Portative vs. Tractive Magnets. Magnets, such as those just 
described, which do work by exerting force through a distance are 
known as tractive electromagnets. There are other electromagnets 
known as portative, or holding magnets which differ considerably 
from tractive magnets. These magnets develop torque but do 
not do any work, simply holding the iron which comes in contact 
with them as long as current is supplied to their windings. It can 
be readily seen that the requirements for this type of magnet are 
altogether different from those for the tractive type. The porta- 
tive magnet requires no moving parts, and the whole problem is 
to get sufficient flux at the proper place. Probably the best-known 
type of portative magnet is the no-voltage release on a motor 
starter. This little magnet merely holds the starting lever in 



261 



22 



ELECTROMAGNETS AND COILS 



position as long as current is supplied to it, and if power fails, the 
magnet loses its magnetism and the starting lever returns to the 
OFF position. 

Effect of Power Supply. Another important factor which 
must be considered in designing a magnet is whether it is to be 
used on an a.c. or d.c. circuit. The same general principles of 
design are used for both, but the design of the a.c. magnet is more 
difficult because of the inductance effects. On alternating current 
the hysteresis and the eddy-current losses also affect the design, and 
it is usually necessary to laminate the iron structure, which makes 
a.c. less satisfactory mechanically as a rule than d.c. apparatus. 

Duty. The duty cycle of the magnet does not affect the type 
of magnet but affects the dimensions. As previously explained, 
if a magnet is only energized one-half the time, the coil will not 
get as hot as it will if the magnet is in circuit all the time. It is 
therefore possible, when a magnet only operates part of the time, 
to increase the power supplied to the coil and so get more work 
out of the magnet. 



DIRECT=CURRENT ELECTROMAGNETS 

Theory. The solenoid-and-plunger electromagnet consists of 
only a coil and a movable plunger. Fig. 11. Although this type 

of magnet is used comparatively little 
in practice, it is essential to know how 
it works in order to explain the char- 
acteristics of the other types. If the 
plunger of the magnet in Fig. 11 be 
placed partly inside the coil and the 
coil energized, the plunger will of course 
be magnetized; mutual attraction will 
then take place and the plunger will 
be drawn into the coil. If the plunger 
is the same length as the coil, it will be 
drawn in [until its ends are even with 
the ends of the coil; but if the plunger 
is longer than the winding, it will be 
attracted until the middle of the plunger is even with the middle 
of the coil. 




Fig. 11. Solenoid and Plunger 



262 



ELECTROMAGNETS AND COILS 



23 



A characteristic pull curve of a solenoid-and-plunger magnet 
in which the coil and the plunger are of equal length is shown in 
Fig. 12. It will be noted that when the plunger is withdrawn 
until the bottom of the solenoid is even with the top of the plunger, 
very little pull is exerted. As the plunger is drawn farther into the 
coil, however, the pull increases until the maximum value is reached. 
The pull is then constant for part of the stroke and drops off to 
zero when the end of the plunger is even with the end of the coil. 



I 

I, 



^ ^ ^ 


/ ^ 


-4 i- 


ao t 


Sc f 


i^ 


4 


^ t 


r .. 


p 


3 


^ t - 


t 


t 


Jz. 


^ t 


X 


t 







^o ^o /aa 



Fig. 12, Pull Curve of Solenoid-and-Plunger Magnet 

The pull at any point is proportional to the product of the 
flux due to the coil and the flux induced in the core, which tends 
to cut across the coil flux. As stated before, the flux within the 
coil is fairly uniform through the middle and drops off at the ends. 
It would therefore be expected that the pull would be less when 
the plunger was just entering the coil. This effect is further in- 
creased by the fact that the core is not saturated until it is well 
within the coil. This explains why the pull is small at the start 
and gradually increases. Through the middle of the coil the excit- 



263 



24 



ELECTROMAGNETS AND COILS 



ing force is constant and the induced flux is practically constant, 
which accounts for the flat part of the pull curve. As the end of 
the stroke is approached, the exciting flux drops off slightly and 
the induced flux remains practically constant, but in this position 
the field due to the coil and that due to the plunger practically 
parallel each other, and the pull therefore drops off until the coil 

and plunger ends are even, when the 

1 ^Do<S^s^s^s^ The point at which the maximum 

pull is reached depends on several fac- 
tors, but it is usually obtained when 
0.4 of the plunger is within the coil. 
By increasing the ampere turns the 
plunger is saturated sooner, and there- 
fore the maximum-pull point is reached 
earlier in the stroke. 

Effect of Iron Stop. The pull of a 
solenoid of the type shown in Fig. 11 
can be increased by placing an iron 
stop at the farther end of the solenoid. 
Fig. 13. This stop has practically no 
effect on the starting pull or on the pull during the early part of the 
stroke, but as the plunger approaches the stop, the pull, instead 
of dropping off as before, now increases and in this way the 
effective range of the solenoid is greatly increased. If the stop is 
made longer so that it extends into the coil, the pull near the end 
of the stroke will be still further increased. The maximum pull 
at the end of the stroke is obtained when the end of the stop is 
near the middle of the coil, but with this position the length 
of stroke is of course reduced, and in practice the stop is usually 
placed near the end of the coil. 





Fig. 13. 



Solenoid and Plunger 
with stop 



FLAT=PLUNQER ELECTROMAGNET 

Return Path of Iron. It is at once apparent that if an iron 
stop increases the pull at the end of the stroke, the pull can be 
still further increased if a complete return path of iron is provided 
for the flux, because in this way the reluctance of the circuit will 
be decreased and the flux increased. For this reason practical 



264 



ELECTROMAGNETS AND COILS 



25 



electromagnets are usually provided with a return path of iron. 
A typical plunger type electromagnet with flat plug and plunger 
is shown in Fig. 14. 

In a long-stroke magnet of the plunger type the pull curve at 
the start of the stroke depends almost entirely on the solenoid 
effect and therefore will be similar to that given for the solenoid- 
and-plunger type, but at the end 
of the stroke the pull will be 
greatly increased. In a short- 
stroke magnet the pull due to the 
solenoid effect is a very small 
portion of the total because the 
reluctance of the air gap is not a 
large proportion of the total 
reluctance. 

Calculation of Pull. In mak- 
ing magnetic calculations it is 
usually only necessary to know 
the maximum pull due to the 
solenoid effect since a magnet is 
seldom worked below this point. 
The maximum pull due to sole- 
noid effect is approximately 





P.= 



Fig. 14. 

INA 



Plunger Type Electromagnet 



(24) 



175 L 

where Ps is the pull in pounds, N is turns, I is amperes, L is 
length of the coil in inches, and A is the plunger area in square 
inches. Equation (24) is only approximate although it is fairly 
accurate for obtaining the maximum pull on long coils, but on 
short coils the ampere turns are considerably higher. The pull due 
to the attraction between the stop and the plunger is 



P -[My A 



(25) 



where Pm is pull in pounds, N, I, and A have the same values 
as in equation (24), and / is the length of the air gap in inches. 
The total pull is therefore 

IN . r NI 



P = Ps + Pm = A 



175 L L2660/ 



]) 



(26) 



265 



26 



ELECTROMAGNETS AND COILS 



This equation is only approximate, and in practice the ampere 
turns are usually increased about 25 per cent to take care of leak- 
age and other factors which have been omitted. Since this equa- 



<:& 




































■ 














































































9^. 




































1 






































1 




































^ 


V 




































y 




V 9^ 
































/ 


j 


































■»> 


// 






> 






























X 


/ 






«; 




























/ 


Ali 


/ 






^/o 


























/ 




7 








t 
























/ 


/ 




/ 








1 






















> 


/ 




/ 
































/ 






/ 






























/ 






/ 










^ 


C 


















/ 


/ 






/ 








-/ 


/ 


\ 
















^ 


/ 






/ 






v^ 


y 


. r / 


/ 


t 
















/ 






/ 


/ 




y 




"V. 


f 


















/ 








/ 


/ 


y 






/ 


•>^ 


■^ 


X 












/ 








/ 


/■ 


^ 






^ 


^ 






1? 










/ 


/ 






/ 


y 






^ 


:/ 




^ 


uii! 


yy. 


>^ 










/ 






/ 


/ 






^ 


/^ 


^ 


I'-* 


^""^ 


<^ 


(o^ 


-4 ^ 








/ 




/■ 


y 


I 


^ 


^ 


,^ 


A 


'^ 


-*^ 


^ 


^ 


:=^ 

\:'^ 




^ 






/ 




^ 


y 


y 


'^ 


^ 


^ 


2-- 


^ 


y^ 


'^ 


^ 


--^ 


\i^ 


> 






/ 


/ 


y 




-:> 


-< 


-^ 


^ 


>=^ 


^ 


^ 




■— 


^ 


I-- 






/ 


/ 


^ 


J 


^ 


^^ 


^ 


*i^ 


^ 


^^ 


^^ 


-^ 


■^ 


i^ 


2 , 




— 




k 


^ 


g 


rn: 


^ 


inr: 


:::! 


— 




— " 


■^ 













Sooo 



So0O 



4000 ifOOO 

Fig. 15. Curve for Calculating Pull of Plunger Electromagnet 

tion is rather difficult to use, it is more convenient to refer to the 
curves in Fig. 15. The pull in pounds per square inch of plunger 
is plotted against ampere turns for different values of L and /. 



266 



ELECTROMAGNETS AND COILS 



27 



Take for instance the curve marked "X = 2." This curve shows 
the pull due to the solenoid effect on a coil 2 inches long. With 
5000 ampere turns the pull is 7 pounds; with 2500 ampere turns, 
3.5 pounds; and so on. 

The pull indicated by the I curves is of course the pull due to 
the attraction between the plug and the plunger. The curve 
marked '7 = |" gives the pull in pounds per square inch at J-inch 
stroke for different values of ampere turns. The pull at 5000 



WL. ^ 1 


^x 


s 


4^^ 1^ 


^' \ 


\ 


^ \ S 


3^ \ ^^^ 


\- \ ^>, 


15 ^^ 


is; ^ 


"ji 1 v \ 


^ ^^ ^«^ 


\ ^v ^^ 


1 !. "^^ ^ 


^ t^^ "V 


'^ ^^ ""^^^""^^ 




"^■^^^ *^s^ 





^2 



/ 



/^ 



^/^ 



Fig. 16. Pull Curve of Plunger Electromagnet 



ampere turns is 8 pounds; at 6000 ampere turns, 11 pounds; and 
so on. 

In plotting the curves in Fig. 15 the ampere turns have been 
increased to take care of leakage and other losses, so that these 
factors can be neglected. 

Example 21. A plunger electromagnet has a coil 4 inches long, and the 
excitation is 5000 ampere turns. The plunger is 1 square inch in cross-section. 
What pull can be obtained at 1-inch stroke? 

Since the coil is 4 inches long, the pull due to solenoid effect at 5000 ampere 
turns will be found from the curve "L = 4," Fig. 15, to be 4 pounds. The 



267 



28 



ELECTROMAGNETS AND COILS 



stroke is 1 inch, and therefore the pull due to the attraction between the plug 
and the plunger at 5000 ampere turns will be found from the curve ''^ = 1" to 
be 2 pounds. The total pull then is 4+2, or 6 pounds. ■ 

Ans. 6 lb. 
Example 22. A plunger electromagnet has a coil 6 inches long, and the 
excitation is 5000 ampere turns. The plunger is 2 square inches in cross- 
section. What pull can be obtained at 2-inch stroke? 

An8. 7 lb. 

Comparison of Theoretical Test Data. In Fig. 16 are shown 
the actual pull curves of the magnet illustrated in Fig. 17. The 




Bj9' ^Lfjoe /too 



CI. r/ea/^e 



3je. Tu3//>f^ 



C.^.I. 7^LU/i^£^ 



Fig. 17. Plunger Electromagnet with Flat Plug 

curve marked B gives the values of pull obtained when the excita- 
tion is approximately 8000 ampere turns. It will be noted from 
the curve that at 1-inch stroke a pull of approximately 22 pounds 
is obtained. Compare this with the values obtained from the 
curves in Fig. 15. The coil is 6 inches long, and the area of the 
plunger 1.92 square inches. At 8000 ampere turns the pull from 



268 



ELECTROMAGNETS AND COILS 



29 



TABLE HI 
Dimensions of Standard Magnets with Flat Plug 



Size 


A 


B 


c 


D 


E 


F 


G 


H 




(in.) 


(in.) 


(in.) 


(in.) 


(in.) 


(in.) 


(in.) 


(in.) 




1 


6f 


lOi 


3 


51 • 


81 


7f 


If 


H 


2&3 


8f 


llf 


31 


7 


101 


8f 


2i 


2i^ 




4 


10 


m 


4 


8 


llf 


9f 


3 


33^ 




5 


llf 


14 


^ 


n 


12i 


10 


3i 


3H 





/^^^ 




CT./T/ 7^LUf^^£JZ 






Illustration Referring to Table III 

the curve ''X = 6" is 4.7 pounds, and the pull from the curve 
'7 = 1" is 5.25 pounds; the total pull per square inch is approxi- 
mately 10 pounds. The pull for 1.92 square inches is then 19.2 



269 



30 



ELECTROMAGNETS AND COILS 

TABLE IV 
Pull Data for Standard Magnets with Flat Plugs 



Size 


Approximate 

Watts 


Pull (lb.) 


i-Inch Stroke 


A-Inch Stroke 


1 

2 
3 
4 
5 


130 
150 
190 
250 
300 


100 
300 
400 
750 
1000 


50 
150 
200 
400 
600 



pounds, which is less than actually obtained by the test. At IJ- 
inch stroke, curve B, Fig. 16, shows that the magnet actually 
pulled 13 pounds, while from Fig. 15 the pull is 14.1 pounds, which 
is very close to that obtained by test. At J-inch stroke the magnet 
actually pulled 40.5 pounds, while the value from Fig. 15 is 46.8 
pounds. 

Curve C in Fig. 16 shows the characteristics of the same mag- 
net at 4000 ampere turns excitation. At f-inch stroke the pull 
is 15 pounds, which corresponds very closely to the value obtained 
from Fig. 15. At 1-inch stroke the results are also about the 
same. At IJ-inch stroke, however, the actual pull is only 4 
pounds, while 6 pounds' pull should be obtained according to 
Fig. 15. The reason for the discrepancy is that at IJ-inch stroke 
and the low excitation the plunger is beyond the point at which 
the maximum pull due to solenoid effect is obtained and the pull 
is accordingly reduced. 

Standard Magnets. The dimensions of a line of short-stroke 
magnets manufactured by the Cutler-Hammer Manufacturing 
Company are given in Table III and the pull data in Table IV. 
The pull values given are those obtained with intermittent-duty 
coils, and the pull with continuous-duty coils would, of course, be 
less. The data given will indicate the proportions which have been 
found satisfactory for magnets of this type. 

Example 23. Design a magnet of the type shown in Fig. 17 for contin- 
uous operation on 230 volts direct current. The pull required is 30 pounds 
at 5-inch stroke. 

Assume first for the purpose of making calculations that the plunger is 
H square inches in cross-section and that L equals 6. The pull per square 



270 



m ELECTROMAGNETS AND COILS 31 

30 
inch will then be — — ■, or 20 pounds. The required ampere turns can be found 
1.5 

from Fig. 15. At 7200 ampere turns the pull from the curve "L = 6" is 4.2 

pounds, and from "^ = 5" the pull is 15.8 pounds, making a total of 20 pounds. 

The ampere turns should be increased 20 per cent to take care of coil heating 

and therefore 

iV/ = 7200Xl.2 = 8640a.t. 

From equation (19) 

E 230 

The area of the plunger is 1^ square inches and therefore the plunger diam- 
eter is 

"6 



S^ 



or 1.38 inch 



Allowing j^ inch for the brass plunger guide and I inch for coil insulation, the 

inside diameter of the coil wiU be 1.83 inches. The approximate size of wire 

can now be found because it is known that PaRi equals 0.0267 ohm and Pa 

must be larger than the circumference of the plunger. Therefore Ri must be 

0.0267 

less than , or 0.00615 ohm. By referring to Table II it is found that 

7rXl.38 

the coil must be wound with No. 28 or larger enamel wire. 

Assume that No. 25 enamel wire is to be used. Ri for No. 25 is 0.0027 ohm. 

As PaRi equals 0.0267 ohm 

„ 0.0267 

ikr=^ = -^=3.14in. 
TT 3.14 

r=M- inside diameter =3.14- 1.83 = 1.31 in. 
L = 6 in. (assumed) 

Since from Table II the turns per square inch Na for No. 25 enamel wire are 
2630 

A^ = LXTxAra = 6Xl.31X2630 = 20670 turns 

i2= A^X Pa Xi2/ = 20670X0.0267 = 552 ohms 

E^ 2302 

W= — = =95.5 watts 

R 552 

TFa = (M + r)XLX2.2 = 4.45X6X2.2 = 58.7 watts 

As W exceeds Wa, the length of the coil must be increased. It can be proved 
that 

^ = ^ . (27) 

Wa L2 

where L is the length of coil assumed above and Li is the length to which L 

must be increased so that the new W will equal Wa 

Therefore 

95.5 Li2 

= — and Li =7.66 m. 

58.7 36 



271 



32 ELECTROMAGNETS AND COILS 

The resistance of a coil 7.66 inches long will be 

7.66 
i2 = 552 X—T" = 705 ohms 
6 

Since from Table II the ohms resistance per pound of No. 25 wire is 32.7 

R 705 
Weight of copper = — - = - — — = 21.5 lb. 
Ry^ 32.7 

_ ^2 2302 ^ 
W = --=— — = 75 watts 
R 705 

7.66 

Wa = X 58.7 = 75 watts 

6 

The coil wound with No. 25 enamel wire is too long in comparison with the 
other dimensions. With the same plunger diameter a shorter coil would be 
obtained by using larger wire, but the weight of the copper would be consid- 
erably greater. 

By increasing the plunger area the ampere turns required will be decreased 
and a cheaper magnet produced. Assume now a plunger with 2 square inches 



^. 



cross-section. The plunger diameter will be a / — , or 1.59 inches. The inside 

diameter of the coil will be 1.59 + ^+i, or 2.03 inches. The pull per square 

30 
inch will be — , or 15 pounds. 

Assume that L equals 6 inches. From Fig. 15 it is found that 6100 ampere 
turns are required. Allowing for heating, the coil should produce 6100X1.2, 
or 7320 ampere turns. 

E 230 

Assume that No. 25 enamel wire is used. Therefore Ri equals 0.0027] 
As PaRi equals 0.0314 ohm 
^ 0.0314 

j^=^ = liif .3.71 in. 

TT 3.14 

!r = M-insidediameter = 3.71 -2.03 = 1.68 in. 
L = 6 in. (assumed) 

iV = LxrxiV« = 6Xl. 68X2630 = 26530 turns 
i2= ATX Pa X/^/ = 26530X0.0314 = 833 ohms 
„, ^2 2302 

'^^-R=^=''-'^^''' 
TF« = 5.39X6X2.2 = 71.2 watts 
Wa is greater than W, and the thickness of the coil will therefore have to' be 
decreased to cut down the weight of copper. 

About 30 per cent of the difference between W and Wa, or 2.5 watts, 
will be taken up in decrease of Wa- Then] 

M+T = — ^- = 5.2in. 
6X2.2 



2T2 



I 



ELECTROMAGNETS AND COILS 
TABLE V 



33 



Design 
Number 


Plunger 

Area 
(sq. in.) 


Wire 
Size 


Coil Length 
L (in.) 


Thickness of 

Winding T 

(in.) 


Weight of 
Coppe. 


Watts 


1 
2 
3 

4 
5 
6 

7 


u 

2 
2 

f 

3 


25 
25 

26 
26 
26 
26 
26 


7.66 
6.00 

7.40 
6.12 
7.00 
6.50 
7.00 


1.31 
1.59 
1.01 
.1.18 
1.00 
1.06 
0.985 


21.5 

23.5 

15.3 

18.0 

15.2- 

16.2 + 

15.6 + 


75.0 
68.7 
66.5 
55.9 
67.0 
62.6 
65.0 



and M equals 3.615 iliches, T equals 1.585 inches, and P^ equals 11.35 inches. 



iV = 6 X 1 .585 X 2630 = 25050 turns 
72 = 25050X11.35X0.0027 = 767 ohms 



Weight of copper = 



R, 



767 
32.7 



= 23.5 lb. 



2302 
Tr=— — = 68.7 watts 
767 

The weight of copper is higher in this coil than in the first coil, owing to the 
fact that this coil is much shorter than the first. If a coU is wound with smaller 
wire, the length of coil can be increased and the weight cut do^s-n. By mak- 
ing calculations similar to those above for No. 26 enamel wire it wiU be found 
that L equals 7.4 inches, N equals 24, .500 turns, R equals 796 ohms, W equals 
66.5 watts, and the weight of copper is 15.3 pounds. 

By increasing the plunger area it has been possible to decrease the copper 
from 19.4 jwunds to 15.3 pounds. By still further increasing the plunger area 
it may be possible to cut the weight of copper still more. By making calcula- 
tions similar to those given the results in Table V were obtained. 
Design 5 is probably the best. The proportions are fairly good, the plunger 
ia not too large, and the weight of copper is low. The coil also has more ampere 
turns than actually needed, and the magnet will therefore have this additional 
factor of safety. Knowing the approximate dimensions of the magnet, it is 
now possible to determine the exact dimensions. 

Since the plunger area selected is 2^ square inches, the plunger diam- 
eter will be 1.78 inches; and it is advisable to change this to 1.75 inches, which 
is a standard size of cold rolled steel. Table III shows that for size 1 the 
plunger diameter is If inches and that with this plunger size a No. 14 brass 
tube with an outside diameter of If inches is used; the same size of tubing can 
be used in the present design. The inside diameter of the coil will then be about 
1| inches, and, allowing | inch for insulation, the inside diameter of the wind- 
ing will be 2 1 inches. The other dimensions given for the size 1 magnet can 
also be followed in this case because the prehminary data corresponds almost 
exactly with that size. The dimension F in Table III is 7| inches; at each 
end of the coil § inch should be allowed for a protecting washer and another 
I inch for insulation, so that the actual length of the winding will be 6 J inches. 



273 



34 ELECTROMAGNETS AND COILS 

30 
The pull per square inch will be -— , or 12.5 pounds. From Fig. 15 the 

ampere turns required will be 5500, and, allowing 20 per cent for coil heating, 
the coil should be designed for 6600 ampere turns. 

230 

PaRi= =0.03486 ohm 

6600 

For No. 26 enamel wire Ri equals 0.0034 ohm and 



Pa- 


0.03486 
0.0034 


= 10.25 


in. 


L- 


= 6.875 in. 






M- 


10-25 „ ^„ . 
= = 3.26 m. 





T = 3.26-2.125 = 1.135in. 

iV = 6.875X1.135X3320 = 25900 turns 

72 = 25900X0.03486 = 903 ohms 

2302 
TF=—— = 58.6 watts 

Tf„ = 4.395X6.875X2.2 = 66.48 watts 



I 



As Wa is greater than W, it will be possible to cut down the weight of copper 
by decreasing the thickness of the coil. 

Assume the new value of Wa to be 64.3 watts 

64.3 

M-]-T=—— —- = 4.25 inches 

6.875X2.2 

M — r = 2. 1 25 in. (inside diameter) 

„ 4.25+2.125 „_„. 

M = ' =3.188 in. 

2 

!r = 1.06in. 

Po = M7r = 10.0in. 

iV = 6.875 X 1 .06 X 3320 = 24200 turns 

R = \0 X0.0034 X 24200 = 822 ohms 

_ 2302 

W= =64.3 watts 

822 

Since from Table II the ohms resistance per pound of No. 26 enamel wire is 52 

822 
Weight of copper = — = 15.8 lb. 
52 

The correct coil will therefore be one wound with 24200 turns of No. 26 enamel 
wire with a resistance of 822 ohms. The over-all diameter of the coil will be 
2.125+2T+I (insulation), or 4.5 inches. The magnet frame should be made 
wide enough to leave some clearance for the coil, and the dimension of 5| inches 
given in Table III should be satisfactory. If a cast-steel frame is used, the 
cross-sectional area of the frame should about equal the area of the plunger; 
but if the frame is made of cast iron, its area should be increased. In Table III, 
size 1, the area of the frame is about 60 per cent greater than that of the plunger, 
and these dimensions can be used. The stop should have the same outside 



274 



ELECTROMAGNETS AND COILS 



35 



diameter as the inside of the brass tube and can be fastened in place as shown 
in Fig. 17. It should extend into the solenoid about 3 inches, so that the air 
gap is at the middle or just above the middle of the coil. 

Cone Plunger Electromagnet 
Advantage of Cone Plunger. In Fig. 16 are shown the char- 
acteristic curves of a plunger type electromagnet, and the curve 



I 

I 















































































































> 




^ 














4S 
























f 






'V 


N 






























j 












\ 






























1 














\ 


























J 


















\ 




4o 




















/ 




















N 


















/ 










































/ 






































* 


o/ 


























32 
















/ 






































/ 
















































































































24 






















































_ 






































/ 












^ 




























/ 
















^ 


N. 
















/<^ 








;? 




















<. 


k. 














/ 






























^ 










i. 










*^ 


^ 








^ 




£ 










'^ 




^ 














i 




./ 


/^ 


















X 














S 


J 


f 


/ 


<s 






















\ 


f\ 












1 


/ 




s. 










r 












V 












1/ 


/ 


















■*"" 






















11/ 










































f 




































_ 





/2 / //Z 2 

Fig. 18. Work Curves of Plunger Electromagnet 



2'/2 



in Fig. 18 shows the work in inch pounds which the magnet can 
do at different lengths of stroke. It will be noticed that for the 
particular magnet shown the maximum work on curve Q is ob- 
tained when the stroke is \ inch; if the stroke is made either 



275 



36 



ELECTROMAGNETS AND COILS 



larger or smaller the total work done will be less. For every 
plunger electromagnet there is some point in the stroke at which 
maximum work will be done, although this point, of course, 
varies with different magnets. 

It has been found that the maximum work due to the attrac- 
tion between the plug and the plunger is done when the reluct- 
ance of the air gap is equal to the reluctance of the rest of the 
circuit. It will be seen at once that with a flat plunger this point 
will be reached when the stroke is very small, because the reluct- 



H 


1 1 

-*- 


M 


1 




1 



(^ 



a 



/o v=v/^ I U ' ^r^o/c'£ 



d 



,r/ 



Fig. 19. Flat Plug Magnet with Lever Motion 

ance of air is extremely high compared with that of the iron cir- 
cuit. The maximum work on the magnet under consideration was 
done when the stroke was \ inch, and a pull of 10 pounds was 
obtained. Suppose that the magnet was to be used to lift 2\ 
pounds a distance of 1 inch. In order to use the magnet for this 
purpose, it would be necessary to use a 4 to 1 lever motion, 
Fig. 19. The magnet would do the work, but the lever motion 
is undesirable. 

Another way to accomplish the same result is to change the 
shape of the plunger. Fig. 20. The total work (pull X distance) 



276 



tnat 



ELECTROMAGNETS AND COILS 



37 



tnat is done owing to attraction is the same with this type of 
plunger as with the flat plunger, but in this case the pull is less 
and the stroke is greater. It has been found that the best results 
are obtained when the angle of the cone is 60 degrees, and this 
angle is used in the following discussion. 

Length of Stroke. In order to get the same work out of the 
cone plunger that was obtained from the flat plunger, the air- 
gap reluctance must be kept the same. 
Now according to equation (4) the reluctance 



f^^^ ^ 



is equal to — -. The area A of the cone is equal 
^A 



to 



7rZ2 



-kX- 




and that of the flat plunger is equal to 

Therefore the new area will be equal 

to twice the old, and to keep the reluctance the 
same, the length of the air gap will have to be 
doubled. This length of air gap is perpendic- 
ular to the faces of the cone, and the stroke is 
of course longer than the gap. With a^ 60- 
degree cone the stroke is twice the air gap. 
The stroke of the cone plunger is therefore 
four times that of the flat plunger. 

Pull of Magnet. According to equation 
(25) the pull of a plunger electromagnet due to 
attraction between the plug and the plunger is 

P -(iiyA 

Since with the cone plunger the area of the plunger and the 
length of the air gap are doubled, the puU perpendicular to the 
surface of the cone is one-half of the pull for the flat plug and 
the total available pull parallel to the direction of travel is only 
one-fourth of that for the flat plunger. It will be seen that the 
stroke has been multiphed by four and the pull divided by four. 
Therefore the total work due to attraction between the plug and 
the plunger is the same in both cases. 

In using equation (25) for cone-plunger magnets the correct 
result will be obtained if A is the cross-sectional area of the 



Fig. 20. Cone Plug 
and Plunger 



277 



38 



ELECTROMAGNETS AND COILS 



TABLE VI 





Dimensions of Standard Magnets 


with Cone Plug 




1 


Size 


A 


B 


c 


D 


E 


F 


G 


(in.) 


(in.) 


(in.) 


(in.) 


(in.) 


(in.) 


(in.) 




22 


6f 


7f 


H 


2f 


f 


61 


Ill 


23 


71 


9 


51 


3 




m 


2|f 




24 


n 


10 


7i 


4 


I 


11 


46^4 




25 


121 


131 


8f 


41 


If 


12^ 


4ff 




26 


13^ 


15 


9^ 


5 


2 


14^ 


5f| 


1 




^. /. r^AME 



3^^3S TU3//V<^ 



C.^.I. 7=^Li/r/^s;e 



Illustration Referring to Table VI 



plunger below the cone and / is the perpendicular distance between 
the cone plug and the plunger. 

Cone Plunger vs. Flat Plunger. Curve E in Fig. 16 is the 
pull curve of a magnet exactly the same as the one in curve C 
except that a cone plunger is used, and the pull on the longer 
strokes is greater than with the flat plunger. Curve E in Fig. 18 
shows the work done by this magnet, and it will be noted that 
the maximum work done with a cone plunger is greater than 
that done with a flat plunger. This is due to the fact that with 



I 



278 



ELECTROMAGNETS AND COILS 



39 



TABLE VII 
Pul! Data for Standard Magnets with Cone Plug 



Size 


Weight of 

Plunger 

(lb.) 


Approx- 
imate 
Watts 


Pull (lb.) 


i-Inch 


1-Inch 


IHnch 


2-Inch 


2i-Inch 


3-Inch 


3Mnch 


4-Inch 




Stroke 


Stroke 


Stroke 


Stroke 


Stroke 


Stroke 


Stroke 


Stroke 


21 


3.5 


90 


28 


24 


18 


12 


6 








22 


5.8 


110 


50 


36 


27 


20 


13 


10 






23 


13.7 


130 


90 


80 


70 


50 


36 


18 






24 


21.5 


200 


175 


150 


130 


100 


90 


60 


50 


40 


25 


33.5 


300 


360 


310 


270 


220 


180 


130 


110 


80 


26 


49.0 


340 


500 


420 


370 


310 


270 


240 


200 


170 



a cone plunger the pull due to solenoid effect is utilized to better 
advantage. The maximum work with a flat plunger occurs at 
J-inch stroke, and if the pull due to solenoid effect is 3 pounds, 
the work due to the solenoid effect will be JX3, or J pounds. 
With the cone plunger the maximum work occurs at 1-inch stroke, 
and the work due to solenoid effect is 1X3, or 3 pounds. 

It will therefore be seen that for general purposes it is desir- 
able to use the cone plunger. The flat plunger should only be 
used for very short strokes and for those applications where the 
load increases as the plunger approaches the plug. 

Standard Magnets. The dimensions of a line of standard 
magnets with cone plungers manufactured by the Cutler-Hammer 
Manufacturing Company are given in Table VI, and Table VII 
gives the pull data for intermittent duty. The pull for continuous 
duty will be considerably less. 

Example 2Jt. The plunger of a magnet with cone plug and plunger is 
2 square inches in cross-section. The coil is 6 inches long. What pull can 
be obtained at 2-inch stroke if the excitation is 8000 ampere turns? 

The coil is 6 inches long and the pull due to solenoid effect can be found 
from the curve "L = 6" in Fig. 15. The pull per square inch at 8000 ampere 
turns is 4.7 pounds, and for 2 square inches the pull is 2X4.7, or 9.4 pounds. 
The stroke is 2 inches and therefore the perpendicular distance between the 
cone faces is 1 inch. The pull due to attraction between the plug and the 
plunger can therefore be found from the curve "i! = l." At 8000 ampere turns 
the pull is 5.3 pounds per square inch, and for 2 square inches the pull is 10.6 
pounds. The total pull is therefore 10.6-|-9.4, or 20 pounds. 

Ans. 20 lb. 

Horseshoe Electromagnets 

Characteristics. A horseshoe type of electromagnet is shown 

in Fig. 21. This type of magnet is used for electric bells and 



279 



40 



ELECTROMAGKETS AND COILS 



relays in which the stroke of the magnet is very short and 
the total work to be done is small. It is evident from Fig. 21 
that if the armature is moved very far from the pole pieces, the 
total length of the air gap will be greater than the distance 
between the poles; most of the flux will then pass directly from 
one pole of the magnet to the other without going through the 
armature. For this reason the horseshoe magnet should only be 
used in case the stroke is short. 

Ampere Turns. The pull of the horseshoe magnet is expressed 
by equation (25), which can also be stated with NI as the 
unknown term, thus 



NI = y^^2m0l 



(28) 



It should be remembered in using equation (28) for the horse- 
shoe magnet that the area A is the sum 
of the areas of the two poles and I is 
the sum of the air gaps, or twice the 
distance between the armature and the 
cores. Equation (28) is only approx- 
imate because there is always a great deal 
of leakage in a horseshoe magnet. The 

I I ampere turns should always be made 25 to 
I 50 per cent more than the equation calls 
for, always keeping in mind that the 
leakage is greater for longer air gaps. 




Fig, 21. Horseshoe Type 
Electromagnet 



Example 25. Assume that the cores of the magnet shown in Fig. 21 
each have an area of \ square inch. The air gap is \ inch, and the pull required 
is I pound. How many ampere turns will be required, assuming that 25 per 
cent additional ampere turns are required to take care of leakage? 

The area of each core is \ square inch, therefore A equals | inch* the air 
gap is \ inch therefore, I equals | inch and P^ equals ^ pound. 

Substituting these values in equation (28) we find 



NI = J=2660 X J = 665 a.t. 



but 25 per cent must be added for leakage and therefore 
iV/ = 665X1.25, or 831 a.t. 



Ans. 831 a.t. 



280 



ELECTROMAGNETS AND COILS 41 

■ Example 26. With the conditions the same as in example 25, what would 
the pull be at i^-inch stroke, assuming that the leakage is the same in both 



Ans. 0.22 lb. 

Example 27. Design a horseshoe magnet to lift 2 pounds at ^-inch 
stroke for use on 230 volts direct current, half-time duty. 

In the solution of example 23 several preUminary designs were worked 
out and the best one chosen and worked out in detail. The same procedure 
would have to be followed in designing a horseshoe magnet, but as the calcu- 
lations are all the same, only the final design is given in this case. 

Assume that the diameter of the iron core will be f inch. The area of 

each core will then be — -Xf X|, or 0.4415 square inch, and the area of the two 

cores A will be 0.883 square inch. The inside diameter of the winding will 
be f +i (insulation), or 1 inch. P^ is 2 pounds and I is | inch because I is the 
total length of air gap. 
From equation (28) 



iV/ = ^/^-^X2660X| = 2006 a.t. 



Allowing 50 per cent additional ampere turns for leakage and coil heating 
AT/ = 2006X1.5 = 3009 ' 



E 230 



NI 30.09 

Assume that No. 33 enamel wire will be used; Table II shows that Ri equals 

0.0172 ohm. 

Therefore 

0.0765 

P„ = = 4.435 in. 

" 0.0172 

4.435 
M = = 1.41 in. 

TT 

r=M— inside diameter = 1.41 — 1=0.41 in. 

Assume that L equals 4 inches (two coils each 2 inches long). 

N = LXTxNa = 4: X0.41 X 16200 = 26570 turns 
R=NXPaXRi = 26570 X0.0765 = 2033 ohms 
E^ 2302 

^ = -^ =:;;;:;:; = 26 watts 

R 2033 

As the coil is to be designed for half-time duty, 1.4 watts should be allowed 
for each square inch of cyUndrical surface. 

Tra = (M+T)XLXirXl.4 = 1.82X4X3.14Xl.4 = 32.1 watts 

As Wa exceeds W, the coil diameter can be decreased. 
Assume the new value of Wa to be 30.4 watts. 

M+T = — = 1.723 in."' 

^ 4X3.14X1.4 



281 



42 ELECTROMAGNETS AND COILS 

Therefore M equals 1.362 inch, and T equals 0.362 inch. 

p^=:M7r = 1.362X3.14 = 4.28 in. 

Pa^i = 4.28X0.0172 =0.0735 ohm 

AT = 4X0.362X16200 = 23460 turns 

R = 23460 X0.0735 = 1725 ohms 

2302 

W = =30.5 watts 

1725 

R 1725 
Weight of copper =—- =—— = 1.29 lb. 

Ky, looZ 

W and Wa are now almost equal, and the winding therefore should have 23,460 
turns of No. 33 enamel wire. As there are two coils, each should be wound 
with 11,730 turns, and the resistance of each coil will be 862.5 ohms. 

The outside diameter of each coil will be M-\-T-{-i (insulation), or 
1.723+0.25, which equals 1.973 inches. Allowing for clearance between coils, 
the two cores should be spaced 2| inches center to center. The over-all length 
of each coil is L + | (insulation), or 2j inches. Allowing y6 inch for a protect- 
ing washer on each end and | inch for the core to project beyond the coil, the 
length of each core should be 2| inches. The iron in the yoke and the arma- 
ture should have approximately the same cross-section as the cores. 

Clapper Type Electromagnet 

Characteristics. A modification of the horseshoe magnet is 
the clapper type electromagnet, Fig. 22, which has characteristics 
similar to the horseshoe type but has only one air gap and one 
coil instead of two. The equation for finding the pull is the same 
as that for the horseshoe magnet, and in designing new magnets 
the same formulas and the same allowances should be used as for 
horseshoe magnets. 

The clapper type magnet is used for small relays and also 
for magnet switches, which will be described later. The pull 
curve is similar to that of the plunger type with flat plug and 
plunger. The pull is very small at the start and increases rapidly 
as the clapper approaches the core, and the point of maximum 
work for this type of magnet occurs when the clapper is very 
close to the core. For this reason the clapper type magnet should 
not be used if the load is constant, that is, if it is desired to lift 
a constant weight through a given distance; it can be used to 
advantage, however, when the load increases as the clapper 
approaches the core. 

Effect of Enlarging Plunger. It is possible to increase the 
starting pull of this type of magnet by enlarging the end of the 



282 



ELECTROMAGNETS AND COILS 



43 



plunger. The effect of the enlarged plunger is to cut down the 
reluctance of the air gap and so increase the flux, and as the pull 

/fluxV 

is proportional to ( j , the pull is of course increased. At the 

\area/ 

end of the stroke, however, the effect of the enlarged core is to cut 
down the pull since, when the clapper is very close to the core, 
the flux is practically the same 
whether the core is large or small, 
and therefore the pull at this point 
in the stroke is less when the area is 
larger. It is well to remember that 
the final pull of a magnet can be in- 
creased by making the area of contact 
smaller. The net result of the en- 
larged core is to increase the starting 
pull and to decrease the final pull. 

Example 28. What pull can be obtained 
from the magnet shown in Fig. 22 if the exci- 
tation is 2000 ampere turns and the air gap \ 
inch? Assume 20 per cent leakage. Arts. 6j lb. 




Fig. 22. Clapper Type 
Electromagnet 



Portative, or Holding, Magnets 

Characteristics. The problem of designing a portative, or 
hoi ing, magnet is quite different from that of designing a tractive 
magnet. With the tractive magnet the main thing is to get as 
much work out of it as possible and the final, or sealed, pull is 
usually not very important. With the portative magnet the 
conditions are reversed, and the magnet is so designed that the 
maximum pull is obtained when the armature is in contact with 
the core. The tractive type magnet is usually rather long with 
a comparatively small diameter; the diameter of the portative 
magnet is large compared to its length. 

Flux Density. The pull of a portative magnet is approx- 
imately 



P = 



72130000 72130000 A 



(29) 



As the holding power depends on the square of the flux density, 
cast steel is usually used for the frame because its permeability is 



283 



44 



ELECTROMAGNETS AND COILS 



a great deal higher than that of cast iron. This can be readily 
seen by referring to the f^-d^ curves shown in Fig. 5. With 100 
ampere turns per inch the flux density for cast iron is approximately 
46,000, while at the same point the flux density for cast steel is 
about 104,000. As the pull depends on the square of the flux 
density, there is a very great advantage to be gained by using 
cast steel. 

A sectional view of a simple type of portative magnet is 
shown in Fig. 23. When cast steel is used for a magnet of this 
type, the usual practice is to make the flux density in the area of 
contact about 110,000 maxwells per square inch and the flux 

density in the rest of the iron 
circuit 90,000 maxwells per square 
inch. 

Temperature Rise. The coils 
in the portative magnet are en- 
tirely enclosed, and the heat must 
be transmitted through the iron 
before it can be radiated. For 
this reason coils for completely 
enclosed magnets are not rated 
the same as those which are exposed to the air. It has been 
found that to limit the temperature rise to 50° C, 3J square 
inches of coil surface must be provided for each watt to be dissi- 
pated. The number of square inches given is for continuous 
duty, and for intermittent duty only approximately one-half the 
surface is required. In magnets of this type not only the outside 
cylindrical surface of the coil is counted but also the inside sur- 
face, top and bottom. 

Exam-pie 29. Design a magnet of the type shown in Fig. 23 which will 
support a weight of 10,000 pounds. Assume 110,000 Unes per square inch 
in the area of contact and 90,000 lines per square inch in the rest of the iron. 
Design the coil for continuous duty on 230 volts. 

The area of contact can be calculated from equation (29), which when 
transposed is 

^^7213O00OP ^3,^ 




Fig. 23. Portative Electromagnet 



72130000X10000 
12 100 000 000 



=59.5 8q. in. 



284 



ELECTROMAGNETS AND COILS 45 

This area is, of course, divided between the inner and the outer poles, and the 
area of each will be 29.75 square inches. Consider first the inner pole. The area 
of contact as determined above is 29.75 square inches. The area behind the 
point of contact, however, will have to be larger, as we have assumed its flux 

density to be only 90,000. This area will therefore be X 29.75, or 36.36 

square inches. The diameter of the inner pole will then be a / — '■ , or 6.8 

inches. Since i^ inch must be allowed around the core for insulation, the 
inside diameter of the coil will be 6.8+3^ + 1^, or 7.18 inches. 
Assume the coil thickness x, Fig. 23, to be 1 inch and the coil length y to be 
2 inches for a trial calculation. If x equals 1 

F« = (7.18 + l)7r = 25.7in. 
and d equals approximately 10 inches, and h about 1.1 inches. The length of 
the magnetic circuit is indicated by the dotted line in Fig. 22 and will be roughly 
13 inches. It has been assumed that the flux density in the iron will be 90,000 
maxwells per square inch. Fig. 5 shows that when M is 90,000 for cast steel, 
^ is 42; that is, 42 ampere turns will be required for each inch length of the iron 
circuit. Since the length of the iron circuit is 13 inches, the ampere turns required 
will be 13X42, or 546. In addition to the ampere turns required for forcing 
the flux through the iron circuit, it is necessary to consider the ampere turns 
required for forcing the flux through the air gap. Even if the cores and arma- 
tures are very carefully machined and cleaned, there will be some air gap. 
Now for air M equals 3.19 ^, and as M in the air gap has been assumed to be 
110,000 

110000 

^=— = 34500 a.t. per m. 

3.19 

If the air gap is assumed to be 0.01 inch, the total length of the path in air will 
be 0.02 inch, 0.01 inch at the inner and 0.01 inch at the outer pole. The ampere 
turns required by the air gap will be 34500X0.02, or 690. The total ampere 
turns required will be 690+546, or 1236. This value should of course be in- 
creased 20 per cent to allow for the heating of the coil, and therefore 

A^/ = 1236X1.2 = 1483 a.t. 
Equation (19) 

may be written 



E 



Ri= = =0.00603 ohm^ 

. ^ NlPa 1483X25.7 

As Ri is 0.00603 ohm, the proper size of wire from Table II is found to be No. 28 
enamel, for which Ri equals 0.0054 ohm. 

Ar = LxTxA^a = 2XlX3320 = 6640'turns 

6X25.7 
Tra=-ri;7- = 47.4 watts 

E^ 2302 . 

W = —= = 57.25 watts 

R 0.0054X6640X25.7 



285 



46 ELECTROMAGNETS AND COILS 

As W exceeds Wa, it will be necessary to increase the length of the coil. From 
equation (27) 

57.25 L\ Li2 

47.4 ~U~ 4 

The correct coil will then be 2.2 inches long and 1 inch thick. 

iV = 2.2 X 1 X 3320 = 7304 turns 

i2 = AT XF«Xi?/ = 7304X25.7X0.0054 = 1013 ohms 

2302 

W= = 52.2 watts 

1013 

R 1013 
Weight of copper = — - = — — = 7.72 lb. 
Kjt, 131.2 

The outside diameter of the inner pole shoe has been found to be 6.8 
inches. The inside diameter of the outer pole shoe will be 6.8+2 (coil) + f 
(insulation), or 9.55 inches. The outside diameter can be found because the 
area of the pole shoe is known to be 36.36 square inches, and if the total area 
of a circle at the inside of the pole is added to the area of the pole, the result 
will be the area at the outside diameter d, Fig. 23, or 

7r9.552 7rd2 

——+36.36 = — 
4 4 

from which 

rf = 11.73 in. outside diameter 

, 11.73-9.55 _^ . 
h= = 1.09 m. 

2 

36.36 

= 1.7 m. 



6.8X3.14 

and if an armature is provided, it should have the same thickness. The area 
of the poles in contact with the armature is to be 29.75 square inches, and it 
will therefore be necessary to reduce these areas as shown in Fig. 23. The 
amount to be cut away from each pole is 36.36 — 29.75 square inches, or 6.61 
square inches. The diameter of the recess in the inner pole wiU be 



4 



4X6.61 ^^. 
/ = J = 2.9m. 



The dimension g can be found in the same way as b was calculated and will be 
found to be 0.9 inch. 

h = 2.0S (coil)+f (insulation) + 1 (coil shield) =2.95 in. 
The principal dimensions have all been determined, and to get a good idea 
of the proportions, the student should make a drawing to scale. 

ALTERNATINQ=CURRENT ELECTROMAGNETS 
Theory. The pull which can be obtained from an a.c. 
magnet can be determined from the equations which were devel- 
oped for direct current. If the ampere turns are the same, the 
pull will be approximately the same whether alternating or direct 
current is used. 



286 



ELECTROMAGNETS AND COILS 



47 



^M The calculations for alternating current are, however, entirely 
aifferent because it is a pulsating current and the current in a 
winding is limited, not only by the resistance of the winding, but 
also by the inductance. On direct current the inductance of the 
winding keeps the current from reaching its maximum value 



n 

§. 



> -J 



iC 


































II 




































1 


































1 


































1 






























s 




\ 


































\ 






























A 




1 


I 


































\ 




























7 






A 




































% 


























6 






\ 


\ 
























> 






V 


\ 


\ 


^ 

»^' 


















y 


X 


5 








\ 




^ 


<^ 










___ 


-p^ 


X^ 




>< 










V 












r*^ 


^^ 


^ 






^ 












\ 








.^^ 


X> 






>- 


X^ 














t' 






/ 








^ 


^^ 








k 

V 










N 




y^ 






V^ 


/; 


^ 


^^ 
















X 




b^ 




:?^ 


^ 














2 






/ 


y 


-^^ 


^ 


rX 
,/ 
























/ 


<0j 


^ 




y" 


X 













_■ _ 








I 


/ 


^ 


^ 
^ 






























Y. 


^ 
































/ 












— 





















Fig. 24. 



V4 /z ^/* 

Characteristic Curves of A. C. Plunger Electromagnet 



immediately, but the only result of this is that if the inductance 
is high, the magnet will be slow in acting. On alternating current, 
however, the current pulsates so rapidly that the inductance must 
be considered; in fact, on a.c. magnets the resistance of the coil 
is practically negligible compared with the reactance. 



287 



48 



ELECTROMAGNETS AND COILS 



The inductance of a magnet varies with the position of the 
plunger or armature. As the plunger approaches the stop, the 
inductance increases and the current therefore decreases. This is 
shown in Fig. 24. The current at 1-inch stroke is 6.2 amperes, 
but when the plunger is in contact with the stop, the current is 
only 0.6 ampere. For this reason the coils of an a.c. magnet are 






•.AO 




1 














t 




3 


















/ 








. _. 












/ 




^ f^ 
















1 


f 


























^ 2 














i 


r 






.0 














"/ 








Is 

K 












6 

nJ 


/ 


















/ 


















/ 


/ 






/ 












/ 






%/ 


/ 




i\. 








/ 




-^-< 


'^ 


/ 












/ 




e> 


^ 














y 


^ 


















MF{XW£LL.S T^EfZ ^Q. INCH 

Fig. 25. Iron Loss in No. 29 Gage Laminations 

usually inexpensive, and design is not limited by the heating of 
the coil as on direct current. 

Iron Losses. An a.c. magnet is, however, limited by the 
heating of the iron. The heating of the iron is due to the eddy- 
current and the hysteresis losses in the iron circuit. The eddy 
currents are the circulating currents which are set up in the iron 



288 



ELECTROMAGNETS AND COILS 49 

circuit perpendicular to the flux. These currents can of course 
be reduced by laminating the iron circuit in the direction of the 
flux and nearly all a.c. magnets are made of laminated iron. 
From No. 26 to No. 29 gage is usually used for this purpose. 

The hysteresis loss is the energy which is expended in forcing 
the flux first in one direction, then in another. 

. The temperature of the iron is usually limited to about 70° C. 
A higher temperature would not damage the iron, but the coil is 
usually in contact with the iron, and if the iron temperature 
is too high, the heat is transmitted to the coil and will eventually 
cause a burnout. Fig. 25 shows the total iron loss per pound of 
metal for No. 29 carbon steel laminations, which are usually used 
on a.c. magnets. With this grade and gage of iron a temperature 
of 70° C. is obtained when the flux density is between 80,000 and 
90,000 maxwells per square inch. 

Windings. The windings of an a.c. magnet are calculated in 
the same way as the windings of a transformer, and the equation is 

£ = 4_^i^/ (3.) 

As O equals /3A, equation (31) may also be written 

^= lo^ 



(32) 



where E is the effective voltage of the circuit, /3 is the flux 
density in maxwells per square inch, A is the area in square 
inches, N is the number of turns, and / is the frequency of the 
circuit in cycles per second. 

The pull of an electromagnet depends on the flux density, 
and an inspection of equation (32) will show that the pull of a 
magnet can be increased by increasing the voltage, decreasing the 
turns, or decreasing the frequency. 

SingIe=Pliase Magnets. Since alternating current is a pulsat- 
ing current, the flux produced by it is a pulsating flux. As the 
pull depends on the flux, the pull will also pulsate; that is, it 
will pass through zero twice during each cycle, and the magnet 
will chatter. To overcome this chattering, a.c. magnets are 
provided with a shading coil. This coil is merely one turn of 
wire imbedded in the end of the stop, which produces a local 



289 



50 



ELECTROMAGNETS AND COILS 




Fig. 26. Arrangement of Two-Phase 
Electromagnet 



flux out of phase with the main flux, so that the total flux, and 
therefore the pull in the air gap, never passes through zero. 

Polyphase Magnets. A 
steady pull can be obtained from 
a polyphase magnet by the proper 
arrangement of the plungers or 
clappers. Fig. 26 shows an ar- 
rangement of a two-phase magnet 
which has been used. A, B, C, 
and D, are the plungers, and the 
load is applied at E. With this 
arrangement the pull at E is 
constant and equal to the maxi- 
mum pull exerted by either set 
of plungers, and there is very 
little tendency to chatter if the plungers are lined up correctly. 
The usual arrangement of a three-phase magnet is shown in Fig. 27. 
With this type of magnet the total pull exerted is the same at 

all times. 

It is evident that polyphase 
magnets, although offering some ad- 
vantages over single-phase magnets, 
are more complicated, more expen- 
sive to build, and less universal in 
their application. For this reason 
nearly all a.c. magnets are of the 
single-phase type, and therefore the 
polyphase magnet will not be considered in the following text. 

Plunger Electromagnet 
Characteristics. In Fig. 24 is shown the pull curve of the 
a.c. plunger type electromagnet illustrated in Fig. 28. Since the 
current decreases as the plunger approaches the stop, the ampere 
turns decrease and the pull due to the solenoid effect drops off. 
The pull due to attraction between the stop and the plunger, 
however, increases rapidly as the plunger approaches the stop. The 
total pull, which is the sum of these two pulls, is therefore nearly 
constant for the first half of the stroke and then increases rapidly. 



' — ' J. — ' J — 



Fig. 27. Arrangement of Three- 
Phase Electromagnet 



290 



ELECTROMAGNETS AND COILS 



51 



It will be noted that the work which the magnet can do is 
maximum at maximum stroke, and a.c. magnets of this type are 
usually worked at the greatest possible stroke to take advantage 
of this characteristic. 

Design Data. In order to design a magnet of this type 
accurately the inductance of the circuit must be calculated. This 




Fig. 28. A. C. Plunger Type Electromagnet 

is a very complicated and difficult calculation, and fairly good 
results can be obtained by using the following proportions: 

If a magnet is used at its maximum stroke, 1 square inch of plunger area 
should be provided for every 8 to 10 pounds pull. 

The coil length should be approximately 2.5 times the stroke. 
The stop should extend into the coil about i of its length. 



291 



52 ELECTROMAGNETS AND COILS 

The maximum flux density should be from 80,000 to 90,000 maxwells per 
square inch. 

Example 30. An a.c. plunger magnet has a plunger 1 square inch in cross- 
section, and the flux density is to be 80,000 maxwells per square inch. How 
many turns will the coil have if the magnet is used on 220 volts, 60 cycles? 

Ans. 1032 turns 

Example 31. Find the plunger area, number of turns in winding, and 
length of coil for a magnet to lift 15 pounds at l|-inch stroke on 440 volts, 60 
cycles. Assume that /§ equals 90,000 and that the pull is 10 pounds per square 
inch of plunger area. 

Ans. Plunger area 1.5 sq. in. 
Turns 1222 
Coil length 3.75 in. 

Other Types of Electromagnets 

Horseshoe and Clapper Type Electromagnets. The pull 
characteristics of the horse-shoe and clapper type magnets are 
the same on alternating current as on direct current. The same 
limitations apply as in the case of the plunger type alternating 
current magnets. The puH is all due to attraction, and there- 
fore on alternating current circuits, as well as on direct 
current circuits, this type of magnet should only be used for 
short strokes. 

Induction Motor Type. It is not practical to build a.c. 
plunger or clapper type magnets which will pull much over 50 
pounds because the inrush currents taken by the larger magnets 
are extremely heavy. It can also be found from equation (32) 
that as the area of the plunger is increased, a point will soon be 
reached at which the number of turns will be less than 1. If the 
required pull is greater than it is possible to obtain with the 
plunger magnet, a squirrel-cage induction motor is sometimes 
used, which drives the load through a gear reduction. Such a 
motor is so designed that it can stall without damage and is in 
effect a revolving solenoid. 

Portative, or Holding, Magnets. There has been no develop- 
ment in alternating current magnets which corresponds to the 
lifting magnet on direct current, and about the only good example 
of an alternating current holding-magnet is the no-voltage release 
on circuit breakers, oil switches, and motor starters. A laminated 
structure of either the clapper or plunger type is used for 
this purpose. 



292 



ELECTROMAGNETS AND COILS 



53 



jjw ELECTROMAGNET APPLICATIONS 

\W General Statement. The applications of the electromagnet 
ire so numerous that it is almost impossible to give all of them. 
Electromagnets are used on electric bells, buzzers, and annun- 
ciators; in the telephone and the telegraph and for railway signal 
iapparatus; for protecting electrical circuits, overload relays, under- 
load relays, and voltage relays; for magnetic brakes, solenoid- 
operated valves, and magnetic 
contactors; and for numerous 
other purposes. In the follow- 
ing text a description is given 
of some of the most useful and 
interesting applications of elec- 
tromagnets. 

Magnetic Clutch. The mag- 
netic clutch, Fig. 29, does the 
same work as a mechanical 
clutch but, being electrically 
operated, has a number of ad- 
vantages; for example, it can 
be controlled from a distance, 
which makes it very useful as 
a safety device, especially in 
rubber plants, in which the 
operators are sometimes caught 
in the rolls. The operator is 
able, by pulling a safety switch, 
to de-energize the clutch and 




Fig. 29. Magnetic Clutch 



disconnect the motor from the 



line shaft and accordingly stop the machine in the least 
possible time. Another decided advantage is that the magnetic 
clutch does not have any moving parts which are affected by 
centrifugal force (which in some mechanical clutches tends to 
engage or disengage the clutch), and for this reason the magnetic 
clutch can be used for higher speeds than the mechanical clutch. 
The magnetic clutch can also be used in connection with synchronous 
motors and squirrel-cage motors which are used to drive loads they 
are unable to start; that is, the motor can be brought up to speed 
with the load disconnected and the clutch can then be energized. 



293 



54 



ELECTROMAGNETS AND COILS 



The iron framework A, Fig. 29, is fastened to the end of the 
driving shaft B, and the coil C is placed in a depression in the 
framework provided for it. The current for the coil is brought 
in through the slip rings H. The friction lining D is also fastened 
to the driving member. The armature E is fastened to the end 
of the driven shaft by means of the spring plate F, which is bolted 
to the flange G. When the current is applied to the coil C, 
armature E is attracted and presses against the friction lining D, 
and the two shafts are then coupled together. As soon as the 
current is shut off, armature E is returned to its normal position 
by means of the spring plate and the driven shaft stops. 

3 /? 




Fig. 30. Separator Pulley 

Magnetic Separator Pulley. Another application of the elec- 
tromagnet is the magnetic separator pulley, the construction of 
which is shown in Fig. 30. The pulley shown in the drawing 
has four coils A connected in series. At each end of the pulley 
and between the coils, iron pole pieces B are provided, and the 
coils are wound so that one disc is a north pole, the next a south 
pole, and so on. In this way a magnetic field is produced between 
the discs, and any iron which comes in contact with the pulley 
will be retained in contact. The current is brought into the pul- 
ley by means of the slip rings C. The operation of the separator 
pulley is shown in Fig. 3L From the illustration it will be seen 
that all the material is passed on a belt over the separator pul- 
ley; the iron will tend to stick to the pulley while the other 
substances will not adhere, the iron being thus removed. 



294 



ELECTROMAGNETS AND COILS 



55 



These pulleys are used in many different industries, for exam- 
ple, for separating steel from brass scrap, and steel and iron from 
duminum; in coal, coke, and power plants to remove pick heads. 




Fig. 31. Operation of Separator Pulley 



bammers, etc., before the coal is taken into the crushers; and for 
protecting crushers and other machinery in various industries. 

Lifting Magnet. One of the best-known applications of the 
electromagnet is the lifting magnet. Fig. 32 shows a cross- 




[c) (£; CB) ce: 

Fig. 32. Lifting Magnet Details 



section of such a magnet built by the Cutler-Hammer Manufac- 
turing Company. The magnet consists essentially of a heavy 
cast-steel framework A in which is placed a strap-wound impreg- 



295 



56 



ELECTROMAGNETS AND COILS 



nated coil B. C and D are the inner and the outer removable 
pole shoes, and E is a manganese-steel coil shield which is sealed 
with waterproof packing to prevent the entrance of moisture. 




296 



ELECTROMAGNETS AND COILS 



57 



The current is brought into the magnet through an armored 
cable F, 

When the magnet is energized, a magnet field, of course, is 
set up, and any iron which comes in contact with the bottom of 
the magnet will be retained. Fig. 33 will give an idea of how 
these magnets are usedjn practice. Although they have many 





Fig. 34. Magnetic Contactor 
Courtesy of The Cutler- Hammer Manufac- 
turing Company, Milwaukee, Wisconsin 



Fig. 35. Magnetic Lockout Switch 
Courtesy of The Cutler- Hammer Manufac- 
turing Company, Milwaukee, Wisconsin 



different appUcations, they find their greatest, field of usefulness 
in the steel industry, in which they are employed for handling 
scrap, pig iron, billets, plates, etc. 

Magnet Switches. There are some magnet switches manufac- 
tured using the plunger type electromagnets for their operation, 
but nearly all are of the clapper type, similar to those shown in 
Figs. 34 and 35. Fig. 34 shows a shunt switch for operation on 



297 



58 



ELECTROMAGNETS AND COILS 



direct current. When the coil is energized, the clapper is attracted 
to the core and the current-carrying contact is brought against 

the stationary contact, and in this 
way the electric circuit is com- 
pleted. This type of switch is 
usually provided with a magnetic 
blow-out, which assists in putting 
out the arc. In Fig. 34 the blow- 
out has been moved up out of 
place in order to show the con- 
tacts clearly. 

The type of contactor known 
as the magnetic lockout switch is 
shown in Fig. 35. The upper coil 
tends to close the switch, and the 
lower coil to hold it open. On 
high currents the pull on the lower 

Fig. 36. Time-Element Overload Relay coil is Sufficient tO hold the Switch 

Courtesy of The Cutler- Hammer Manufac- . , „ .. 

turing Company, Milwaukee, Wisconsin Open, DUt aS tUC CUrrCUt " lalls, a 





Fig. 37. Magnetic Brake 
Courtesy of The Cutler- Hammer Manufacturing Company, Milwaukee, Wisconsin 



298 



ELECTROMAGNETS AND COILS 



59 



point is reached where the upper, or closing, coil overcomes the 
pull of the lockout coil, and the switch closes. The current at 
which the switch operates can be adjusted by moving the adjust- 
able disc at the bottom of the clapper. This type of switch is 
used for the automatic acceleration of d.c. motors. 

Relays. There are so many different types of relays made for 
various purposes that it is impossible to describe them all. Fig. 36 
shows an overload relay which can be used to protect either a.c. 
or d.c. circuits. The current at which this relay will trip can be 
adjusted by moving the weight on the calibrating lever. When 
an overload occurs, the plunger is of course drawn down into the 
coil. At the bottom of the 
relay there is an oil dashpot, 
which retards the motion of 
the plunger; with this ar- 
rangement the time required 
for the relay to trip depends 
on the amount of the over- 
load — ^the greater the over- 
load, the quicker the circuit 
will be opened and the load 
disconnected. 

Magnetic Brakes. A 
magnetic brake for use on 
direct current is shown in 
Fig. 37. The construction of the magnet on this brake differs 
from most designs in that the two halves of the magnet are 
pivoted about the pin in the base of the brake frame. 

A brake for use on a.c. circuits is shown in Fig. 38. This 
type is provided with a small a.c. motor, which drives the brake 
rods through a gear reduction. 

FAULTS IN ELECTROMAGNETS 
Residual Magnetism. One of the commonest troubles with 
electromagnets is that, because of residual magnetism in the iron, 
they do not let go when the coil is de-energized. This difficulty 
can be remedied by putting a nonmagnetic spacer (usually brass) 
in the iron circuit. 




Fig. 38. Motor-Released Brake for Use on 

Alternating Current 

Courtesy of The Cutler- Hammer Manufacturing 

Company, Milwaukee, Wisconsin 



299 



GO ELECTROMAGXETS A^^D COILS 



Cofl Troaiples. Faflme ctf coils is doe to aevml cubes, but 
is one of tiie moal: cannon. Cbik heS ofvokat and 
faani out if tliey do not have sulBiiru 
for idle CBOgT to be dbsqnted. Fang cofls on 
linn thej ivere designed for oAen icsdIIs in fdne. As die vatts 
as the sqpme of tlie \%jibagt, a smal increase in Toltige 
a rompaialii^y lus^ increase m Iwaiii i g. When a coO 
bums out orving to oiobeating, it is usaa%^ ncoessanr to 
it cntirdy becanse ibc insnlatinn b ccoBM s diaiicd and the 
^enoalhr oocms in the niddle of tbe vinfing, ^dddi is, of conacL 
tbe hottest part. 

The short-cbcnitii^ of tmiB inside the mifing is anodicr 
frequent canse of oita h ea ling. S the ^vltage bda m n tmns i^ 
ha^ or if the la\ns aie not wimnd erenhr, the vollaee beini a tii 
two tmns may be h^ mnof^ to break down tiie ■—"*■*■«* and 
^Mct-cncnit the intcrv^enb^ tune. Hieciectof this is to decrease 
die cofl Rsisfanoev and as the waits are inii a ady proportional to 
die resbtancCp the heating is increased. The presence of —natni c 
decreases the resistanoe stil more, and when co3s are to be nsed 
in damp hwations, ^nimjl pranision ishowM be made to heep 
mobtmcoot of the windn^s. 

Frequenthr coib short-cncmt Ofwii^ to npropcr insolation 
of the inside lead where it is brongjht oat aver the end of the 
winding and b connected to the coil tenmnl Cofls which fiul 
in diis war can osnalhr be repaired, because die bmnuut most 
f ic q ua id% <Mnns near the outside of the winfing. 

fffgilation breakdowns in coib geneialbr occur when the cofl 
^^^-z^ ;^ bn^cn becaiee a U^ i^olt^e k indnrrd in die coiL 
7 jced \i^fage is usuaEjr seraal tones as great as the 

essed Tvitage and depends on the ^leed of open i n g^ the 
the iiniiit is €4irnedj the h^jher the 

INDUCTION COILS 



coib, as thej are usualy caAed, consist of a cofl of 
on an inn core made 19 of a bundle of 9oft4ran vi 
ooib are used in dectnc^gv i^tiag sjrstenB and l: : 



ELECTROMAGNETS AND COILS 



61 



ri 



and gasoline-engine ignition. The coils are simply connected in 

series with the make-and-break device on a 6-to-12-volt battery 

circuit, Fig. 39. When the circuit 

is completed, the current in the coil 

does not immediately rise to its final 

value because the flow of current is 

opposed by the voltage induced in 

the winding by the changing flux. 

Fig. 40 shows how the current builds 

up in a coil of this kind, and it will 

be seen that an appreciable time 

elapses before the current finally 



nmiiiinm 



Fig. 39. 



Wiring Diagram of Primary 
Induction Coil 



E 



rises to its maximum value, which is — (Ohm's law). 

XL 



Fig. 40 also 



shows that when the circuit is broken, the current dies down much 
faster than it builds up. 

When the circuit of the coil is broken, the flux in the iron 
circuit dies down and an e.m.f. is again induced in the winding. 
The value of the induced e.m.f. at the break depends upon the 
speed with which the circuit is broken, because any induced 
voltage depends upon the rate at which the lines of force are cut. 
When the flux is reduced to zero in 0.01 second, the induced 
voltage is twice as much as when the flux is reduced to zero in 
0.02 second. For this reason the contact mechanism is arranged 
so that the circuit is broken with a snap, and an induced voltage 




lNT£X/iSU/''ro/Z OP£/z/^T£a 



Fig. 40. 



77A7i£ 

Building Up of Current in Primary Coil 



equal to 4 to 10 times the battery voltage is usually obtained. 
As a result of this high induced e.m.f., a bright hot spark is pro- 



301 



62 



ELECTROMAGNETS AND COILS 



duced at the point of rupture. This spark is used to ignite 
the charge in a gas [engine or to Hght the gas in a gas Hghting 
system. 

Secondary Induction Coils. Secondary induction coils have 
in addition to the primary coil a secondary coil wound over the 
primary coil. The secondary coil is wound with a great many 
turns of fine wire, and as the induced voltage depends on the 
turns, a very high voltage is induced in the secondary coil when 
the primary is broken. 

Uses. The first successful induction coils were built about 
1840, but for about fifty years they had very little practical 

^ Q_ value and were used only for 

experimental purposes. When 
X-rays were discovered in 
'^ 1890, however, the induction 
J coil became a commercial 
device as it was found that 
the discharges from a second- 
ary coil provided the best 
means of operating the X-ray 
tubes. Static machines can 
also be used for this purpose 
but their polarity frequently 
reverses, which causes dete- 
rioration of the platinum 
plates which are used as the anodes of the tubes. 

Secondary induction coils are also used in wireless telegraphy, 
and when so used, an operating key is substituted for the auto- 
matic interrupter. 

Another application of secondary coils is for gas-engine igni- 
tion of the jump-spark type. In some ignition systems a separate 
induction coil is used for each spark plug, and each coil is pro- 
vided with its own interrupter. In other systems a separate coil 
is used for each spark plug, but a master interrupter or vibrator 
is used for breaking all the circuits. In still other systems only 
one induction coil is used, and this is connected successively to 
the different spark plugs by means of a distributor. With this 
system the energy is usually supplied by a high-tension magneto. 




Fig. 41. Wiring Diagram of Secondary 
Induction Coil 



302 



I ELECTROMAGNETS AND COILS 63 

, Coils used for medical purposes are of the same general 
esign as other secondary coils, but are arranged so that the cur- 
rent strength may be regulated. This is usually accomplished 
by sliding a metal tube on or off the core, thus regulating the 
time in which the flux dies down and accordingly regulating the 
voltage induced in the secondary coil. 

Operation. The connections for a secondary induction coil 
are shown in Fig. 4L When the switch J is closed, the circuit 
is completed to the primary coil F through the contact mechanism 
at E and current flows through the primary coil. The core A is 
then magnetized, and the clapper Z), the end of which is made 
of iron, is attracted to the core. When the clapper D moves 
toward the core, the circuit to the primary coil is broken at E, 
the flux of course dies down, and a high voltage is induced in the 
secondary winding K. When the circuit of the primary coil is 
broken at E, sparking will occur just as it does on a spark coil 
when the circuit is broken; as a result the flux decreases more 
slowly and therefore the voltage of the secondary coil is cut down. 
For this reason the condenser C is shunted across the interrupter 
terminals, and when the circuit is broken at E, the energy of the 
primary coil is not dissipated in an arc but charges the condenser 
C instead. In this way the primary circuit is broken more 
quickly which very materially increases the voltage of the second- 
ary coil. L is the spark gap across which the secondary coil 
discharges. 

After the flux has died down in the core A, the clapper D 
is released and returns to its original position, and the process 
outlined above is repeated, the coil continuing to operate as long 
as the switch J is left closed. 

Rating. Secondary coils are usually rated according to the 
length of spark produced. The length of spark, however, does 
not completely define a coil since the energy of two coils of equal 
spark length may be entirely different. For X-ray work high 
voltages with comparatively little energy have been found most 
suitable, but for wireless work a shorter spark and a larger cur- 
rent have proved to be most satisfactory; for this reason larger 
wire is used in winding the secondaries of coils designed for 
wireless work. 



303 



64 



ELECTROMAGNETS AND COILS 






CONSTRUCTION OF INDUCTION COIL 
Core. The cores of induction coils are made up of bundles 
of soft annealed iron wires. A bundle of wires is used in place 
of a solid core on account of the eddy currents which are set up 
when the flux is changed rapidly. The dimensions of the core 
should be carefully selected; if the core is too small in proportion 
to the other dimensions, the total flux, and consequently the 
induced voltage of the secondary, will be low. It has been found 
that as the iron in the coil is increased, the secondary voltage of 
the coil is increased. With a large core the reluctance is less and 
it is possible to make and break the primary circuit faster than 
with a smaller core. On the other hand, increasing the core 
diameter beyond a certain point does not give a proportional 



^£cor^r>/^RY Co/i- 



CO^£ 




f/^T£J5ISUF^£C 



B/iTT£^r 



Fig. 42. Cross-Section of Small Induction Coil • 

increase in secondary voltage. Increasing the core diameter also 
increases the average length of turn of the coils and accordingly 
increases the weight of copper and the cost. It is therefore 
apparent that unless too large a core is avoided, the cost of the 
coil will be excessive. Proportions for coils to produce sparks 
of 2, 6, and 12 inches are given in Table IX. The core is usually 
insulated- from the primary winding by being wrapped with 
manila paper, or the wires can be put into an insulating tube. 

Primary Coil. The primary coil usually consists of two layers 
of comparatively heavy wire which are wound over the core 
insulation. Either silk- or cotton-covered magnet wire is used for 
this purpose, and as the voltage between turns in the primary is 
not high, cotton-covered wdre, which is cheaper, is usually used. 
The size of primary wire depends on the size of the coil but 



804 



ELECTROMAGNETS AND COILS 



65 



should be chosen so that the current density in the wire is not 
over 600 to 1200 ampere turns per square inch, because the 
primary coil has no ventilation and is liable to overheat. 

For coils for short spark lengths spool heads are fastened to 
each end of the core and both the primary and the secondary 
coils are wound in the space thus provided, insulation of course 
being provided between the two coils, Fig. 42. For larger coils, 
however, the primary is wound on the iron core and an insulating 
tube slipped over it; the secondary is then wound separately and 



} 


S 

7 

3 

2 
/ 






1 














/ 


















/ 


/ 
















/ 


/ 


















/ 






1 














/ 


















/ 


/ 








^i 










y 


/ 


















y 


/ 
















y 


/ 














^ 























Fij 


5.43. 


Spar 


AT/ 

king 


DistaE 


ce be 


tween 


s 

Need 


e Poi 


tits 



put in place on the insulating tube. The size and amount of 
wire required for various coils is given in Table IX. 

Secondary Coil. The number of turns required in the second- 
ary coil to produce a given spark depends on three factors: the 
value of flux in the core; the speed with which the primary 
circuit is broken; and the position of the secondary coil. The 
position of the secondary coil is important because the magnetic 
field is strongest at the middle of the core, and as it is important 
to get the winding in the strongest part of the field, it is usually 
about one-half as long as the core. The strength of the magnetic 



805 



66 



ELECTROMAGNETS AND COILS 



field also decreases as the distance from the core increases, and 
for this reason the outer turns of a coil are not so effective ^s 
those closer to the core. 

The curve in Fig. 43 shows the voltage necessary to produce 
a spark between needle points at different distances. It will be 
noted from the curve that for a 2-inch spark a voltage of approx- 
imately 35,000 is necessary, and that for a 6-inch gap the voltage 
must be 70,000. It will readily be appreciated that extreme care 
must be taken in insulating the secondary coil or breakdown of 
insulation will result. For spark gaps larger than 2 inches the 
secondary is wound in sections, owing to the high voltage which 

would exist between turns if a 
continuous coil were used. These 
sections are insulated from each 
other and connected in series. 
The coils are generally wound 
with silk-covered or d.c.c. wire, 
which is passed through a bath 
of paraffin while being wound. 
On account of the high volt- 
age in the secondary, great care 
should be taken in constructing 
these coils to see that no moisture 
is present, and the coils are 
usually impregnated with par- 
affin. Ebonite or hard rubber is ordinarily used to insulate the 
secondary from the primary. 

Interrupters. There are a great many different types of 
interrupters in common use, the one selected depending on the 
size of coil, the voltage of the supply circuit, and the service. 
The simplest kind of interrupter is the one shown in Fig. 42. It 
cannot be easily adjusted for different speeds and is in general 
slower than any of the other types. It is satisfactory, however, 
for small spark coils and medical coils. 

In Fig. 44 is shown the atonic interrupter, which is the type 
in general use for ignition and for the larger and more powerful 
induction coils. The clapper A is retarded by the spring 5, and 
the initial position of the interrupter can be regulated by the 




Fig. 44. Atonic Interrupter 



306 



ELECTROMAGNETS AND COILS 



67 



adjusting screw C. The contact mechanism consists of the lever D 
and the adjusting screw E. The advantage of this interrupter Ues in 
the fact that when the clapper A is attracted to the core, it moves 
some distance before it touches the lever B and therefore strikes 
it with a sharp blow. The effect is to open the circuit much 
quicker than with the simple type of interrupter. Another advan- 
tage of the atonic interrupter is that it can be set for a differ- 
ent speed of vibration by adjusting the spring tension and the 
initial position of the clapper. 

The electrolytic interrupter is another type which is used to 
some extent. The Wehnelt interrupter consists of a lead-plate 

I 

I 

^ .2 .^ .6 .S ^ /, 

Fig. 45. Effect of Capacity on Spark Length . 

cathode and an anode of platinum wire immersed in a dilute 
sulphuric acid solution. When current flows through the inter- 
rupter, gas is given off at the anode, which increases the resist- 
ance at that point; and as the current increases, more gas is 
given off until finally the circuit is practically interrupted. 
The current at which the circuit is interrupted can be adjusted 
by moving the platinum wire up and down. One of the advan- 
tages of this type of interrupter is that a condenser is not nec- 
essary. 

There are two types of mercury interrupters which are used 
for high-speed work. In one type a contact point is alternately 







^ 


^^ 
















I 


/ 






\ 


\ 












A 


/ 








s 


\ 










1 


/ 












^^ 


-^ 







I 






















1 






















-f 


1 











































307 



68 ELECTROMAGNETS AND COILS 

dipped into and withdrawn from a well of mercury, and the' 
circuit is broken each time the point is withdrawn. In the other 
type a stream of mercury plays on a revolving wheel resembling 
a sprocket; the circuit is made when the stream of mercury comes 
in contact with a tooth and broken when it passes between two 
teeth. 

Condensers. The condenser is an extremely important part 
of the induction coil, and unless a condenser of the proper capacity 
is used, satisfactory results will not be obtained. The curve in 
Fig. 45 shows the effect of varying the capacity of the condenser. 
The curve shows that with a certain value of capacity the spark 
length is a maximum, and if the capacity is either increased or 

decreased, the spark length is less. 
77// /^0/jL v^^ It is difficult to predetermine 

^^^£^ ^N^ the capacity necessary for any 
given coil because it depends on 
the source of primary e.m.f., the 
primary coil, and the interrupter. 
The most satisfactory way is to 
find the most effective capacity 
by experiment. For large coils it 

Fig. 46. Plate Type Condenser IS advisablc tO prOvidc a COUdenSCr 

of adjustable capacity. 
Plate type condensers are built up of alternate layers of 
metal foil (generally tin foil) and paper, mica, or some other 
dielectric, Fig. 46. For condensers of small capacity thoroughly 
dried paper can be used as the dielectric; but for the larger sizes 
mica or paper which has been carefully treated with paraffin or 
some like substance should be used. The capacity of a condenser 
is directly proportional to the area of the plates and to the 
number of plates. It is inversely proportional to the thickness 
of the dielectric, and therefore the dielectric should be made 
as thin as possible and- still not break down when voltage is 
applied to the condenser. The capacity of a condenser also 
depends on the material used for the dielectric. For equal thick- 
ness of dielectric the capacity of a mica condenser is from two 
to two and one-half times that of a condenser made up of paraf- 
fined paper. 




ELECTROMAGNETS AND COILS 

TABLE VIII 
Specific Inductive Capacities 



69 



ffl 



-t 

Dielectric 


Specific Inductive 
Capacity, k 


Ail- 
Mica 

Paraffined paper 
Rosin 
Glass 
Beeswax 
Shellac 


1 

5 to 8 

2 to 3 
1.86 

3 to 10 
1.95 

3 to 3.5 



Calculations. If the capacity required is known, the dimen- 
sions of the condenser can be calculated from the following 
equation: 

aXnX2.24:Xk 



C = 



txw 



(33) 



where C is the capacity in microfarads, a is the area of the metal 
sheets, n is the number of sheets of the dielectric material, t is 
the thickness of the sheets, and k is a. constant which is called 
the specific inductive capacity. The specific inductive capacity 
of a dielectric is the ratio of the capacity of a condenser made of 
the dielectric to the capacity of a condenser of the same dimen- 
sions using air as a dielectric. The value of k for different 
dielectrics is given in Table VIII. 

Example 32. Find the capacity of a condenser made up of 300 sheets of 
paraffined paper. The tinfoil sheets are 4 inches long by 5 inches wide, and the 
thickness of the paper is 0.002 inch. Assume that k equals 2.5. 

Ans. 1.68 microfarads 

Design of Secondary Induction Coils. There are no hard 
and fast rules which can be stated for the design of induction 
coils. Nearly all coils are the result of individual experiments, 
and new coils are usually designed by the cut and try method. 
The data given in Table IX represents average practice and will 
give some idea of the proportions. 

Tesla Coil. The Tesla coil is the type of coil used by Nikola 
Tesla in his high-voltage high-frequency investigations. The coil 
has a primary winding of few turns and a secondary winding of a 
comparatively larger number of turns but is not provided with an 



309 



70 FT-ECTROMAGNETS AND 


COILS 




TABLE IX 






Induction Coil Dimensions 




Part 


2-Inch 


6-Inch 


12-Inch 


Spark 


Spark 


Spark 


Core 








Size iron wire 


22 


22 


22 


Diameter in inches 


u 


If 


2i 


Length in inches 


10 


14 


24 


Insulation between core and primary 








coil in inches 


1^ 


i 


1 

4 


Primary coil 








Size wire B.& S. 


14 


12 


10 


Turns 


200 


225 


330 


Layers 


2 


2 


. 2 


Ebonite tube between primary and 








secondary 








Inside diameter in inches 


1| 


2h 


31 


Outside diameter in inches 


2 


3 


4| 


Secondary coil for X-ray, etc. 








Wire size B.& S. 


38 


36 


34 


Approximate weight of wire in 








pounds 


3 


8 


15 


Secondary coil for wireless 








Wire size B.&S. 


34 


32 


30 


Approximate weight of wire in 








pounds 


6.8 


18 


33 


Condenser capacity in microfarads 


1 


2 


5 



^r/^i/^K <^/7/=» 




iron core. The Tesla coil is energized from the secondary of a 
secondary induction coil, and even higher voltages are obtained 

on a Tesla coil than on a sec- 
ondary coil. Even greater care 
must be taken with the insula- 
tion of a Tesla coil than is neces- 
sary with that of a secondary 
induction coil. 

Several different ways of 
connecting these coils have been 
used, one of which is shown in 
Fig. 47. The discharge from the 
secondary coil I energizes the 
primary coil P of the Tesla coil 
and charges the condenser C, and when the voltage induced in the 
coil 1 dies dcv/n, the condenser C discharges. The discharge of a 
condenser is an. oscillating discharge of high frequency, and the 
current in the coil P is accordingly a high-frequency current. 



fflsim 



Fig. 47. Wiring Diagram of Tesla Coil 



310 



ELECTROMAGNETS AND COILS 71 

Every time the induction coil I operates, it causes a number of 
pulsations in the coil P, and the frequency of the Tesla coil is a 
great many times that of the induction coil. The frequency of 
the secondary coil S is the same as that of the coil P, but the 
voltage induced in S is higher than that in P owing to its larger 
number of turns. The voltages induced in the coil S are there- 
fore of extremely high voltage and frequency. 



I 



311 



V. 



Mi 



I 



i; 



INDEX 



I 



The 'page numbers of this volume will he found at the bottom of the pages; 
the numbers at the top refer only to the section. 



Page 



Air gap 50 

All-day efficiency 169 

Alternating current, relation to 

pressur3 199 

Alternating-current electromagnets 286 
iron losses 288 

other types 292 

plunger type 290 

polyi)hase magnets 290 

single-phase magnets 289 

theory 286 

windings 289 

Armature winding 2, 10 

Y^o"^P' series motor 43, 47 

yVhp. 12-volt shunt motor 37, 40 
yVhp. 110-volt shunt motor 29, 35 
|-hp. 1 10-volt compound motor 55, 58 
automobile starter 117,124,126 

charging generator 131 

rewound motors 85, 90, 93,95, 96, 98 
SDA frame 52 

Asbestos-covered wire 251 

Atonic interrupter 306 

Automatic-regulator farm light- 
ing outfits 137 
Automobile starting and lighting 



equipment 


113 


charging generator design 


126 


classification 


113 


starting motor design 


113 


Autotransformers 


225 


B 




Back-e.m.f. 




|-hp. induction motor 


62 


5-hp. induction motor 


75 


Bearing loss 




T^Trhp- series motor 


44, 46 



Page 

Bearing loss (continued) 

yVhp- 12-volt shunt motor 38, 39 

yVhp. 110-volt shunt motor 32, 34 

i-hp. induction motor 68, 71 

|-hp. 110-volt compound motor 56,58 

|-hp, induction motor 80, 81 

automobile starter 124 

Bearings 8 

Belt-driven farm lighting outfits 130 

Bobbin-wound coils 251 

Booster autotransformer 225 

Breakaway torque 121 

Brush losses 

y^y-hp. series motor 44, 46 

yVhp. 12-volt shunt motor 38, 39 

yVhp. 110-volt shunt motor 32, 34 

|-hp. induction motor 69, 71 

|-hp. 1 10-volt compound motor 56, 58 

^-hp. induction motor 80, 81 

automobile starter 123, 124 

Brush rigging 8 

C 
Case for transformer, design of 231 
case for oil insulation 232 

uses of oil 231 

Charging generator design 126 

classification 126 

combined starting motor and 

charging generator 133 

third-brush generator 130 

vibrator-controlled generator 133 
Circular mils 230 

Clapper type electromagnet 282, 292 
Closed-core transformers 164 

Coefficient of inductance 160 

Coils 

of electromagnet 1S5, 188, 190 

construction of 249 

bobin-wound coils 251 



Note. — For page numbers see foot of pages. 



313 



INDEX 



Page 
Coils (continued) 
construction of 

cost 251 

form-wound coils 252 

general considerations 249 

kinds of wire 250 

winding calculation 253, 289 

troubles in 300 

of transformer 185, 188, 190 

Commutator 6, 52 

Comparison between small and 

large d.c. motor 10 

Compound motor 52 

Condensers 308 

Cone-plunger electromagnet 275 

advantage of 275 

vs. flat-plunger type 278 

length of stroke 277 

pull of magnet 277 

standard magnets 279 

Constants of design 99 

Copper losses 

in motors 

1 o'^P- series motor 45, 46 

•y^hp. 12-volt shunt motor 39 
Y3-hp. 110- volt shunt motor 34 
|-hp. induction motor 69, 70, 71 
^-hp. 110-volt compound 

motor 54, 55, 58 

^-hp. induction motor 81 

automobile starter 124 

in transformers 165, 234, 236, 238 
10-volt transformer 
22-volt transformer 
2200-volt transformer 
10000-volt transformer 
Core 

of induction coils 
of transformers 

10-volt transformer 
22-volt transformer 
2200-volt transformer 
10000-volt transformer 
Core loss (see Iron losses) 
Core transformers 164 

Counter-electromotive force 155 

Crusher autotransformer 226 

Note. — For page numbers see foot of pages. 



490 

503 

202, 212 

172 

304 



215 

227 
204 
179 



173, 



Current, electric 
Cylindrical coils 



Page 
146 
253 



D 



Design constants 99 

Design of electromagnets and in- 
duction coils (see Elec- 
tromagnets and induc- 
tion coils, design of) 
Design of small motors (see 
Motors, design of small) 
Design of smaU low- and high- 
tension transformers (see 
Transformers, design of 
smaU low- and high- 
tension) 
Differential compound generator 126 
Direct-connected farm hghting 

outfits 141 

Direct-current electromagnets 262 

clapper type 282 

cone-plunger type 275 

flat-plunger type 264 

horseshoe type 279 

portative, or holding, type 283 

theory 262 

Direct-current motors 2 

changing characteristics by re- 
winding (see Rewinding 
motors, changing char- 
acteristics by) 
description of 2 

armature 2 

bearings and brush rigging 8 

commutator 6 

comparison between small 

and large motor 10 

field structure 6 

design of 23 

Y^Q-hp. series motor 40 

yg-hp- 12-volt shunt motor 36 
T^hp. 110-volt shunt motor 23 
j-hp. motor 52 

|-hp. 110-volt compound 

motor 52 

design of type SDA frames 323 

Direction of rotation, rewinding 

motor to change 99 



314 



INDEX 



Double cotton-covered wire 



E 



Page 
250 



Eddy-current losses (see Iron 



Efficiency of transformers 168, 

198, 214, 239 
all-day efficiency 169 
definition 168 
financial 239 
ratio of transformation 170 
regulation of transformer 170 
summary 171 
variation in 169 
Electric current, relation to mag- 
netic field 146 
Electrolytic interrupter 307 
Electromagnets and induction 

coils, design of 241-311 
electromagnets 241 
applications of 293 
coil construction 249 
coil design 256, 257 
faults in 299 
theory of magnetic circuit 241 
types of 241, 261 
a.c. electromagnets 286 
d.c. electromagnets 262 
winding calculation 253, 289 
induction coils 300 
construction 304 
condensers 308 
core 304 
design of secondary induc- 
tion coils 309 
interrupters 306 
primary coil 304 
secondary coil 305 
Tesla coil 309 
primary induction coils 300 
secondary induction coils 302 
Electromotive force 29, 150 
coefficient of inductance 160 
counter-electromotive force 155 
effect of number of turns in 

primary 165 

Note. — For page numbers see foot of pages. 



Page 

Electromotive force (continued) 
effect of number of turns in 

secondary 165 

fundamental equation of in- 
duced pressure 152 
induced current 150 
left-hand rule 155 
magnetic leakage 163 
modified equation of induced 

pressure 153 

open- and closed-core trans- 
formers 164 
operation of transformer 161 
rate of change in current 159 
rate of change in magnetic flux 157 
relative motion between circuit 
and magnetic field nec- 
essary 154 
results of moving sHder to right 151 
right-hand rule 151 
shell and core types 164 
voltage equation 29 
Enamel wire' 251 
F 



Farm lighting outfits 


135 


belt-driven systems 


136 


description 


135 


direct-connected sets 


141 


size of outfit 


142 


voltage regulation 


135 


Field coUs 


6 


Y^Q-hp. series motor 


45,47 



T^hp. 12- volt shunt motor 36, 40 
Y^5-hp. 1 10-volt shunt motor 26, 35 
|-hp. 1 10-volt compound motor 

53,58 
automobile starter 120, 126 

charging generator 131 

rewound motors 88, 90, 91, 

95, 96 
Flat-plunger electromagnet 264 

calculation of pull 265 

comparison of theoretical test 

data 268 

vs. cone-plunger type 278 

return path of iron 264 

standard magnets 270 



315 



INDEX 






Page 

Flux (see Magnetic flux) 

Form-wound coils 252 

Frame of motor 22 

y^-hp. 110-volt shunt motor 23 

design of induction-motor frames 83 

design of type SDA frames 47 

air gap 50 

heating 50 

magnetic circuit 47 

mechanical design 51 

output equation 50 

Frequency 

importance in transformer de- 
sign 194, 238 
rewinding motor to change 105 



Gap density ^ 106, 112 

Generator, charging (see Charg- 
ing, generator design) 

H 

Half-speed conditions of series 

motor 41, 46 

Hand-regulated farm Ughting 

outfit 139 

Hand starting 13 

Heating 

allowance for in electromagnets 260 
of motor frame 50 

Hehx 247 

Holding electromagnets (see Por- 
tative electromagnets) 
Horseshoe electromagnets 279 

a.c. type 292 

d.c. t\i3e 279 

ampere turns 280 

characteristics 279 

Hysteresis loss (see Iron losses) 

I 

Induced current 150 

Induced pressure, equations for 

152, 153 
Inductance 

coefficient of 160 

unit of 161 

A'd^e. — For page numbers see foot of pages. 



Page 
Induction, electromotive force of 
(see Electromotive force 
of induction) 150 

Induction coils (see Electromag- 
nets and induction coils, 
design of) 
Induction motor type 292 

Induction motors 12, 59 

characteristics 59 

description 

comparison between smaU 

and large motor 18 

methods of starting 13 

design of 59 

i-hp. motor 59 

|-hp. motor 73 

|-hp. motor 73 

|-hp, 25-cycle motor 83 

frame of motor 83 

rewinding (see Rewinding 
motors, changing char- 
acteristics by) 
Insulation of transformer 188, 231 

Interrupters 306 

Iron, effect of in magnetic circuit 248 
Iron losses 

iQ electromagnets 288 

in motors 

lOQ -hp. series motor 44, 46 

yVhp- 12-volt shunt motor 38, 39 
y^-hp. 110-volt shunt motor 

31,34 
|-hp. induction motor 68, 7l 

|-hp. 110-volt compound 

motor 56, 58 

|-hp. induction motor 79, 81 

automobile starter 123, 124 

in transformers 166, 234 

10-volt transformer 490 

22-volt transformer 503 

2200-volt transformer 202, 213 
10000-volt transformer 172, 173 
Iron stop, effect of in electro- 
magnet 264 



Leakage Unes 



164 



316 



INDEX 



Left-hand rule 155 

Lifting magnet 295 

Losses (see Bearing loss, Brush 
losses, Copper losses, 
Iron losses. Stray-power 
loss, and Windage loss) 
Losses of transformer, propor- 
tioning 172, 214, 227 

M 

Magnet switches 297 

Magnetic brakes 299 

Magnetic circuit 47, 241 

coil construction 249 

effect of iron in magnetic cir- 
cuit 248 
flux 243 
flux density 246 
law of magnetic circuit 241 
magnetic intensity 246 
magnetomotive force 242 
permeability 244 
relations of B & H 246 
reluctance 243 
reluctance of magnetic paths 

in parallel 245 

reluctance of magnetic paths 

in series 245 

simple magnetic circuit 247 

winding calculation 253 

Magnetic clutch 293 

Magnetic conductivity 148 

Magnetic field, relation to elec- 
tric current 146 
Magnetic flux 148, 243 
density of 236, 246, 283 
distribution of 

j^Q-hp. series motor 40 

Y3-hp. 12-volt shunt motor 
1 10- volt shunt motor 



T5-hp. 

|-hp. induction motor 

|-hp. induction motor 

rewound motors 
rate of change in 
Magnetic induction 
Magnetic intensity 
Magnetic leakage 

Note. — For page numbers see foot of pages. 



,46 
36 
24 
60 
73 
100 
157 
150 
246 
163 



Page 

Magnetic separator pulley 294 
Magnetizing current 

|-hp. induction motor 65 

^-hp. induction motor 76 

rewound motors 102, 108 

Magnetomotive force 242 

Main winding 

|-hp. induction motor 64 
§-hp. induction motor 75 
rewound motors 101, 106 
Mean gap density 112 
Mechanical design of motor frame 51 
Mercury interrupters 307 
Mercury-tube regulator 128 
Motors, design of small 1-143 
automobile starting and light- 
ing equipment 113 
changing characteristics of 

motor by rewinding 85 

farm-lighting outfits 135 

general principles 1 

illustrative designs 22 

d.c. motor design 23 

induction motor design 59 

N 

No-load primary loss 238 

No-load speed of series motor 42 

O 

Oil insulation 189, 231 

Open-core transformers 164 

Output equation for motor frame 50 



Paraffin-dipped transformer coils 185 
Performance of motors, compari- 
sons between small and 
large motors 12, 20 

Permeability 148, 244 

Plunger, effect of enlarging 282 

Plunger electromagnet 262, 

264, 275, 290 
characteristics 290 

design data 291 

Poles, number of as affecting 

speed of motor 111 



317 



INDEX 





Page 


Polyphase magnets 


290 


Portative electromagnets 


261, 283 


a.c. type 


292 


d.c. type 


283 


characteristics 


283 


flux density 


283 


temperature rise 


284 



vs. tractive electromagnets 261 

Power efficiency of transformers 
(see Efficiency of trans- 
formers) 
Power factor 201 

Pressure 
. relation to alternating current 199 
relation to winding and con- 
necting transformer coils 

190, 223, 224 
Primary coil of induction coil 304 

Primary induction coils 300 

Primary loss (see also Copper 



Primary winding of transformer 165 
10-volt transformer 213, 217 

22-volt transformer 227 

2200-volt transformer 198, 207 

10000-volt transformer 175, 181 

Proportion, computing new wind- 
ing by, using old wind- 
ing as basis 112 

Protected electromagnet coUs 260 

Pull of electromagnet 26.5, 277 



R 



192 



Radiation from windings 
Rate of change 

in current 159 

in magnetic flux 157 
Rating 

of motors 10, 18, 89, 95, 104, 110 

of secondary induction coils 303 

Ratio of transformation 170, 213 

Reactance voltage 58 

Regulation of transformer 170 

Relays 299 

Reluctance 243 

of magnetic paths in parallel 245 

of magnetic paths in series 245 

Note. — For page numbers see foot of pages. 



Page 
Repulsion induction motor 14 

Residual magnetism 299 

Resistance loss (see Copper losses) 
Retentivity 149 

Return path of iron in electro- 
magnet 264 
Rewinding motors, changing 

characteristics by 85 

d.c. motors 85 

case I 85 

case II 90 

case III — change of voltage 96 
case IV — change of speed 97 

case V — change in direction 

of rotation 99 

induction motors 99 

case VI 100 

case VII — change of fre- 
quency 105 
case VIII — change of vol- 
tage 110 
case IX — change of speed 111 
Right-hand rule 151 
Right-hand screw rule 147 
Rotation, rewinding motor to 

change direction of 99 

Rotor ' 20, 100 



Secondary coil of induction coil 305 
Secondary induction coils .302, 309 
Secondary loss (see Copper losses) 
Secondary winding 
of motor 

|-hp. induction motor 67 

f-hp. induction motor 79 

rewound motors 104, 110 

of transformer 165 

10-volt transformer 221 

22-volt transformer 229 

2200-volt transformer 203, 209 
10000-volt transformer 178, 184 
Series field of |-hp. 1 10-volt 

compound motor 54 

Series motor 40 

Series winding of rewound motor 93 
Shell transformers 164 



318 



Page 

53 

36 

23 

92 

251 

250 

289 



flux 
dis- 



electromag- 
262, 

motor 



L „ „ 

^H compound motor 

^Pnimt motor 
»^^hp. 12-volt 
^ ^hp. no-volt 

Shunt mnding of rewound motor 
Silk-covered wire 
Single cotton-covered wire 
Single-phase magnets 
Sinusoidal distribution of 
(see Magnetic flux, 
tribution of) 
SHp 

Solenoid-and-plunger 
net 
, Space factor 

Speed, rewinding 

change 
Split-phase starting 
Square-core coils 
Square mils 
• Standard magnets 
Standardized frames 
Starting motor design 
low-voltages employed 
requirements 
starter for engine of 200 cubic 

inches displacement 
starter for engine of 300 cubic 
inches displacement 
'* starting motor and charging 

generator combined 
Starting winding 

|-hp. induction motor 
|-hp. induction motor 
rewound motors 
Stator 

Step-down transformers 
10-volt transformer 
22-volt transformer 
characteristics of 
Step-up transformers 

2200-volt transformer 197 

10000-volt transformer 171 

Stray-power loss 31, 38, 46, 58 

' Stroke of cone plunger, length 

I of 277 

Ir Note. — fvr page numbers see foot of pages. 



INDEX 



Page 



70 

290 
29 
to 

97, 111 

13 

260 

230 

270, 279 

22 

113 

115 

113 



124 



117 



133 

67 

77 

103, 109 

20, 100 

213 
226 
191 



Tables 

comparison of small and large 

d.c. motor 9 
comparison of small and large 

induction motor 19 
data applying copper magnet 

wire 177 
dimensions of standard mag- 
nets with cone plug 278 
dimensions of standard mag- 
nets with flat plug 269, 270 
enamel wire table 255 
induction coil dimensions 310 
pull data for standard magnets 

with cone plugs 279 

single cotton-covered wire table 254 

specific inductive capacities 309 

table of equivalents 153 

Temperature rise in portative 

electromagnet 284 
Tesla coil 309 
Third-brush generator 127, 131 
Tractive electromagnets vs. por- 
tative electromagnets 261 
Transformer oil 189, 231 
Transformers, design of small 

low- and high-tension 145 

definition 146 

practical design 171 

10-volt transformer 213 

22-volt transformer 226 

2200-volt transformer 197 

10000-volt transformer 171 
design of case for 220-volt 

transformer 231 
miscellaneous details 190 
recapitulation 234 
regulation of 170 
theory of transformer operation 146 
electromotive force of induc- 
tion 150 
elementary princii^les 146 
use of 145 



U 



Universal motors 



21 



319 



INDEX 



Vibrating contact regulator 
Vibrator-controlled generator 
Voltage change in motors 

rewinding to secure 96, 

without rewinding 
Voltage equation 
Voltage ratio 

Voltage regulation of farm light- 
ing outfits 



Page 

129 
133 

110 

111 

29 

110 

135 



Windage loss (continued) 

i-hp. induction motor 69, 71 

|-hp. 110-volt compound motor 

57, 58 



307 



W 

Wehnelt interrupter 

Windage loss 

Y^o^-hp. series motor 44, 46 

yVhp- 12-volt shunt motor 38, 39 
yV^P- 110-volt shunt motor 33, 34 

Note. — For page numbers see foot of pages. 



|-hp. induction motor 
Winding calculation for electro- 
magnet 
a.c. t5T)e 
d.c. type 

allowance for heating 
cylindrical coils 
design suggestions 
procedure in designing coil 
protected coils 
size of wire 
square-core coils 
Wire, kinds of 



81 



289 
253 
260 
253 
257 
256 
260 
256 
260 
250 



^y 



0619 



320 






LIBRARY OF CONGRESS 

021 213 140 4 



ijikftl 






ill 



i H 



